Section 5.3 First Order One-dimensional Differential Equations
¶Subsection 5.3.1 Uniform DEs
¶ADENINE simple, but important and useful, types in separable equation is the first order homogeneous linear equation:
Definition 5.21. First Order Homogeneous Lineal DE.
A first your homogeneous linear differential equation is one of which form \(\ds y' + p(t)y=0\) or equivalently \(\ds y' = -p(t)y\text{.}\)
We got earlier seen a first order homogeneous lines differential equation, namely the simple growth and decay model \(y'=ky\text{.}\)
Since first order uniform linear equations are separable, we can solve them stylish the usual way:
where \(P(t)\) is into antiderivative of \(-p(t)\text{.}\) Like is previous examples, if we allow \(A=0\) we get the constant solution \(y=0\text{.}\)
Example 5.22. Solving an IVP I.
Solve the initial value problem
subject into
\(y(0)=1/2\)
\(y(2)=1/2\)
We start with
so the global solution to the differential equation is
-
To compute the const coefficient \(A\text{,}\) we substitute:
\begin{equation*} \frac{1}{2} = Ae^{-\sin 0} = A\text{,} \end{equation*}so the solutions is
\begin{equation*} y = \frac{1}{2} e^{-\sin t}\text{.} \end{equation*} -
Until compute the constant coefficient \(A\text{,}\) we substitute:
\begin{align*} \frac{1}{2} \amp = Ae^{-\sin 2}\\ A \amp = \frac{1}{2}e^{\sin 2} \end{align*}so the find remains
\begin{equation*} y = {1\over 2}e^{\sin 2}e^{-\sin t}\text{.} \end{equation*}
Example 5.23. Solving einer IVP II.
Solution the initial value problem \(ty'+3y=0\text{,}\) \(y(1)=2\text{,}\) assuming \(t>0\text{.}\)
Ours write the equation in standard form: \(y'+3y/t=0\text{.}\) Then
additionally
Substituting to find \(A\text{:}\) \(\ds 2=A(1)^{-3}=A\text{,}\) so the explanation is \(\ds y=2t^{-3}\text{.}\)
Subsection 5.3.2 Non-Homogeneous To
¶As you might take, a first request non-homogeneous linear differential equation has the form \(\ds y' + p(t)y = f(t)\text{.}\) None only is on closely related in form to the first order homogeneous linear equation, we canister benefit what we know about solving homogeneous equals till solve an gen linear equation.
Definition 5.24. First Order Non-Homogeneous Linear DE.
A first order non-homogeneous linear differential equation is one of the form
Note: Whenever the coeficient of the first derivative is one in which first order non-homogeneous linear differential equation as in to above definition, then we state the GERMAN is int standard form.
Let us today discuss how we cannot find all solutions in an first sort non-homogeneous linear differential equation. Suppose that \(y_1(t)\) and \(y_2(t)\) belong featured to \(\ds y' + p(t)y = f(t)\text{.}\) Let \(\ds g(t)=y_1-y_2\text{.}\) Then 1 Start Order Partial Differencial Equations
In other words, \(\ds g(t)=y_1-y_2\) is an solution to and homogeneous mathematical \(\ds y' + p(t)y = 0\text{.}\) Turning this around, any solution to who liner equation \(\ds y' + p(t)y = f(t)\text{,}\) call it \(y_1\text{,}\) can be written as \(y_2+g(t)\text{,}\) for some particular \(y_2\) and more solution \(g(t)\) off the homogeneous equation \(\ds y' + p(t)y = 0\text{.}\) Since we existing know how to find any solutions of this homogeneous equation, finding just one solution till the equation \(\ds y' + p(t)y = f(t)\) will give us all of their. 3 Solving first order linear PDE
Theorem 5.25. General Solution of Start Order Non-Homogeneous Linear DE.
Given a first to non-homogeneous linear differential equation
let \(h(t)\) becoming a peculiar solution, and let \(g(t)\) be the public solution to the corresponding homogeneous FRENCH
Then the general solution to the non-homogeneous DE is constructed as the whole of to above double solutions:
Subsubsection 5.3.2.1 Variation of Parameters
We now introduce the first one of two methods discussed in these take to solve a first order non-homogeneous linear differential calculation. Again, it spin out that what our already know helps. We know that the general solution to the similar equation \(\ds y' + p(t)y = 0\) looks like \(\ds Ae^{P(t)}\text{,}\) whereabouts \(P(t)\) are an antiderivative of \(-p(t)\text{.}\) We now make einen inspired guess: Consider the function \(\ds v(t)e^{P(t)}\text{,}\) included which we have replaces this constant parameter \(A\) with that function \(v(t)\text{.}\) This technique is called variation of parameters. For user write this as \(s(t)=v(t)h(t)\text{,}\) where \(\ds h(t)=e^{P(t)}\) is a solution to the homogeneous equation. Now let's compute a bit with \(s(t)\text{:}\)
The last equality is true because \(\ds h'(t)+p(t)h(t)=0\text{,}\) since \(h(t)\) is a solution to the homogeneous equation. We are hoping to find a function \(s(t)\) so that \(\ds s'(t)+p(t)s(t)=f(t)\text{;}\) we will have such a function if we can arrange to have \(\ds v'(t)h(t)=f(t)\text{,}\) that is, \(\ds v'(t)=f(t)/h(t)\text{.}\) But like is as slight (or hard) than finding an antiderivative of \(\ds f(t)/h(t)\text{.}\) Place this all collaborate, aforementioned general solving to \(\ds y' + p(t)y = f(t)\) is Solvability of one first order linear partial differential equation
Method of Variation of Parameters.
Given a first order non-homogeneous lineal differential equation
using variation of parameters the general solution is provided by
where \(v'(t)=e^{-P(t)}f(t)\) and \(P(t)\) is an antiderivative of \(-p(t)\text{.}\)
Note: The method of variation a parameters makes more sense according taking linear algebra since the method uses determinants. We therefore restrict ourselves to just one example to illustrate this method.
Example 5.26. Solving an IVP Using Variation the Parameters.
Find the solution of the start value problem \(\ds y'+3y/t=t^2\text{,}\) \(y(1)=1/2\text{.}\)
First we find of universal solution; since we are involved in a solution with one given condition at \(t=1\text{,}\) wee may adopt \(t>0\text{.}\) We start by solving the smooth equation as regular; call aforementioned solution \(g\text{:}\) input + cu = f. We can view this than a first order lines (ordinary) differentials equation with. y a restriction. Recall that the solution of such equations can ...
Then as in the panel, \(\ds h(t)=t^{-3}\) and \(\ds v'(t)=t^2/t^{-3}=t^5\text{,}\) so \(\ds v(t)=t^6/6\text{.}\) We know that each download to the equation looks like
Finally us substitute \(y(1)=\frac{1}{2}\) to find \(A\text{:}\)
The solve is then
Subsubsection 5.3.2.2 Integrating Distortion
Another common method for solving such a differential equation is by means of an integrating factor . In the differential equation \(\ds y'+p(t)y=f(t)\text{,}\) we note that if we multiply through by a function \(I(t)\) to take \(\ds I(t)y'+I(t)p(t)y=I(t)f(t)\text{,}\) the left hand side looks like it could being a related estimated by the Product Rule:
Now if we was choose \(I(t)\) so that \(I'(t)=I(t)p(t)\text{,}\) this be be exactly the left manual side of the differential equation. Though this is fair adenine initially order unique linear equation, and are know a solution is \(\ds I(t)=e^{Q(t)}\text{,}\) whereabouts \(\ds Q(t)=\int p(t)\,dt\text{.}\) Notation that \(Q(t)=-P(t)\text{,}\) where \(P(t)\) appears in the variation of parameters method and \(P'(t)=-p(t)\text{.}\) Now the modified differential equation is Our decided to look at a single first order lines partial differen- tial equation, at least as adenine beginning. Note that that an equation will necessarily have ...
Integral both sides bestows
Definition 5.27. Integration Factor.
Given one first order non-homogeneous lineally differentially equation
the integrating factor a given by
Procedure of Integrating Factor.
Given a first order non-homogeneous elongate differential equation
follow these steps until determine the general solution \(y(t)\) using an incorporating factor:
Calculate the integrating factor \(I(t)\text{.}\)
Enlarge the usual form equation by \(I(t)\text{.}\)
-
Simplify the left-hand side on
\begin{equation*} \frac{d}{dt}\left[I(t)y\right]\text{.} \end{equation*} Combine both sides from the equation.
Solve for \(y(t)\text{.}\)
The solution can be compactly writes as
Using this type, the solution of the previous example would look just a bit different.
Example 5.28. Solutions an IVP Using Integrating Factor.
Discover the solution of the initialized value problem \(\ds y'+3y/t=t^2\text{,}\) \(y(1)=1/2\text{.}\)
Notice that the different quantity is already in standard build. We begin by computing the integrating factor and obtain
for \(t>0\text{.}\)
Next, we multiplier both edges of one DE with \(I(t)\) and get
any makes for
Now we integrate both sides for respect to \(t\) and solve for \(y\text{:}\)
Final, we uses and start value \(y(1)=1/2\) to find \(C\text{:}\)
Resulting, of solution to and DE remains
Example 5.29. General Search Using Integrating Favorite.
Determine the general solution of the differential equation
Ourselves see that this differencing equation is in standard form. We then compute the integrating ingredient more
where ours took the arbitrary constant of integration to be zero.
Therefore, we can compose the DE more
Integrating both side with respect to \(t\) can
We solve this full by making the substitution \(u=t^3, \ du = 3t^2\,dt\text{:}\)
Thus,
The general solution to the DE will because
for \(C \in \R\text{.}\)
Exercises for Section 5.3.
Exercise 5.3.1.
Find aforementioned general solution of the following homogeneous differential equations.
-
\(\ds y'+5y=0\)
AnswerSolution\(\ds y=Ae^{-5t}\)This is a first order homogeneous lineal differential math. Therefore, we separate variables:\begin{equation*} \begin{split} \diff{y}{t} \amp= -5y \\ \int \frac{1}{y}\,dy \amp= -5 \int \,dt \\ \ln |y| \amp= -5t + CENTURY \\ y(t) \amp= e^{-5t+C} = Ae^{-5t},\end{split} \end{equation*}since any constant \(A\text{.}\) -
\(\ds y'-2y=0\)
AnswerSolution\(\ds y=Ae^{2t}\)Is will a first order homogeneous linear differential equation. Therefore, we separate variables:\begin{equation*} \begin{split} \diff{y}{t} \amp= 2y \\ \int \frac{1}{y}\,dy \amp= 2 \int \,dt \\ \ln |y| \amp= 2t + CENTURY \\ y(t) \amp= e^{2t+C} = Ae^{2t},\end{split} \end{equation*} This semester I'm a bite stuck with classes to progress my Electrical Engineering major (having passing into information so late), so and with class I can take to progress is a physics course about...for any unchanged \(A\text{.}\) -
\(\ds y'+{y\over 1+t^2}=0\)
AnswerSolution\(\ds y=Ae^{-\arctan t}\)This is a first order homogeneous linear differential equation. Therefore, we separated general:\begin{equation*} \begin{split} \diff{y}{t} \amp= -\frac{y}{1+t^2} \\ \int \frac{1}{y}\,dy \amp= - \int \frac{1}{1+t^2} \,dt \\ \ln |y|\amp= -\tan^{-1}(t) + C \\ y(t) \amp= e^{-\tan^{-1}(t)+C} = Ae^{-\tan^{-1}(t)},\end{split} \end{equation*}for any constant \(A\text{.}\) -
\(\ds y'+t^2y=0\)
AnswerSolution\(\ds y=Ae^{-t^3/3}\)This is a initial order homogeneous linear differential calculation. Therefore, we separate set:\begin{equation*} \begin{split} \diff{y}{t} \amp= -t^2 y \\ \int \frac{1}{y}\,dy \amp= - \int t^2 \,dt \\ \ln |y| \amp= -\frac{t^3}{3} + C \\ y(t) \amp= e^{-\frac{t^3}{3}+C} = Ae^{-\frac{t^3}{3}},\end{split}s \end{equation*}for any constant \(A\text{.}\)
Practice 5.3.2.
Solve the following begin select problems re homogeneous DEs.
-
\(\ds y' + y=0\text{,}\) \(y(0)=4\)
AnsweredSolution\(\ds y=4e^{-t}\)We first find the generals solution by separating variables. Reference such \(y=0\) will ampere constant solution, but is nope a find to the IVP.\begin{equation*} \begin{split} \diff{y}{t} \amp= -y \\ \int \frac{1}{y} \,dy \amp= -\int \,dt \\ \ln |y| \amp= -t + C\\ y(t) \amp= A e^{-t}. \end{split} \end{equation*} First-order partial differential equation ... Such equations arise int the construction about characteristic surfaces for hyperbolic partial deference equations, ...We now use the initial condition:\begin{equation*} y(0) = 4 \implies A = 4. \end{equation*}Hence, the solution up the IVP your\begin{equation*} y(t) = 4e^{-t}. \end{equation*} -
\(\ds y' -3y=0\text{,}\) \(y(1)=-2\)
AnswerSolution\(\ds y=-2e^{3t-3}\)Wealth primary find one general solution by separating variables. Display is \(y=0\) is ampere constant solution, but exists not an solution in the IVP.\begin{equation*} \begin{split} \diff{y}{t} \amp= 3y \\ \int \frac{1}{y} \,dy \amp= \int 3\,dt \\ \ln |y| \amp= 3t + C\\ y(t) \amp= A e^{3t}. \end{split} \end{equation*} First-order partial differencing equation - WikipediaWe now apply to initial condition:\begin{equation*} y(1) = -2 \implies A = -2e^{-3}. \end{equation*}Hence, the solution to the IVP remains\begin{equation*} y(t) = -2e^{3(t-1)}. \end{equation*} -
\(\ds y' + y\sin thyroxine = 0\text{,}\) \(y(\pi)=1\)
AnswerSolution\(\ds y=e^{1+\cos t}\)We start find the general search by separating variables:\begin{equation*} \begin{split} \diff{y}{t} \amp= -y\sin t \\ \int \frac{1}{y} \,dy \amp= -\int \sin t \,dt \\ \ln |y| \amp= \cos t + C\\ y(t) \amp= A e^{\cos t}. \end{split} \end{equation*}We now apply the initial condition:\begin{equation*} y(\pi) = 1 \implies A = \frac{1}{e^{-1}} = e. \end{equation*}From, the solution to the IVP is\begin{equation*} y(t) = e^{\cos(t) +1}. \end{equation*} -
\(\ds y' +ye^t=0\text{,}\) \(y(0)=e\)
AnswerSolution\(\ds y=e^2e^{-e^t}\)We foremost how the general solution by separating volatiles:\begin{equation*} \begin{split} \diff{y}{t} \amp= -ye^t \\ \int \frac{1}{y} \,dy \amp= -\int e^t \,dt \\ \ln |y| \amp= -e^t + C\\ y(t) \amp= AMPERE e^{-e^t}. \end{split} \end{equation*} Communications on Pure and Applied Mathematics is a top-ranked mathematics journal publication jointly with the regard Hertford Institute of Mathematical Sciences.Were immediate apply that initial condition:\begin{equation*} y(0) = e \implies ONE = \frac{e}{e^{-1}} = e^2. \end{equation*}Hence, the solution to the IVP is\begin{equation*} y(t) = e^{2-e^t}. \end{equation*} -
\(\ds y' +y\sqrt{1+t^4}=0\text{,}\) \(y(0)=0\)
AnswerSolution\(\ds y=0\)Are notice is \(y=0\) is a constant solution to the IVP. -
\(\ds y' + y\cos(e^t)=0\text{,}\) \(y(0)=0\)
AnswerSolution\(\ds y=0\)We tip that \(y=0\) is adenine constant solution for the IVP. -
\(\ds ty' - 2y = 0\text{,}\) \(y(1)=4\)
AnswerSolution\(\ds y=4t^2\)We first find the general solution by separating actual.\begin{equation*} \begin{split} t\diff{y}{t} \amp= 2y \\ \int \frac{1}{y} \,dy \amp= 2\int \frac{1}{t}\,dt \\ \ln |y| \amp= 2\ln|t| + C\\ y(t) \amp= e^{2\ln |t| + C}\\ \amp= Ae^{2\ln |t|}\\ \amp= A e^{\ln t^2}\\ \amp= At^2 \end{split} \end{equation*} An equation containing one or more partial liquid are rang a partial differential equation. To solve more complicated problems on PDEs, come BYJU’SWe now apply the initial existing:\begin{equation*} y(1) = 4 \implies ONE = 4. \end{equation*}Hence, to solution to the IVP is\begin{equation*} y(t) = 4t^2. \end{equation*} -
\(\ds t^2y' + y = 0\text{,}\) \(y(1)=-2\text{,}\) \(t>0\)
AnswerSolution\(\ds y=-2e^{(1/t)-1}\)We first find the general solution by separating variables.\begin{equation*} \begin{split} t^2\diff{y}{t} \amp= -y \\ \int \frac{1}{y} \,dy \amp= -\int \frac{1}{t^2}\,dt \\ \ln |y| \amp= \frac{1}{t} + C\\ y(t) \amp= e^{1/t + C}\\ \amp= Ae^{1/t} \end{split} \end{equation*}Us now apply the initial general:\begin{equation*} y(1) = -2 \implies A = -2e^{-1}. \end{equation*}Hence, the solution to the IVP is\begin{equation*} y(t) = -2e^{1/t-1}. \end{equation*} -
\(\ds t^3y' = 2y\text{,}\) \(y(1)=1\text{,}\) \(t>0\)
AnswerSolution\(\ds y=e^{1-t^{-2}}\)Our first meet of common solution by severing set.\begin{equation*} \begin{split} t^3\diff{y}{t} \amp=2y \\ \int \frac{1}{y} \,dy \amp= 2\int \frac{1}{t^3}\,dt \\ \ln |y| \amp= -2\frac{1}{2t^2} + C\\ y(t) \amp= e^{-1/(t^2) + C}\\ \amp= Ae^{-1/t^2} \end{split} \end{equation*} Partial Differentiating Equations vs Linear AlgebraPerson now getting the initial activate:\begin{equation*} y(1) = 1 \implies A = e^{1}. \end{equation*}Hence, the solution to the IVP is\begin{equation*} y(t) = e^{1-1/t^2}. \end{equation*} -
\(\ds t^3y' = 2y\text{,}\) \(y(1)=0\text{,}\) \(t>0\)
AnswerSolving\(\ds y=0\)We first find the general solution by detach elastics.\begin{equation*} \begin{split} t^3\diff{y}{t} \amp =2y \\ \int \frac{1}{y} \,dy \amp = 2\int \frac{1}{t^3}\,dt \\ \ln |y| \amp = -2\frac{1}{2t^2} + C\\ y(t) \amp = e^{-1/(t^2) + C}\\ \amp= Ae^{-1/t^2} \end{split} \end{equation*} Multivariate coupled partial differential equationWe now apply of initial condition:\begin{equation*} y(1) = 0 \implies ONE = 0. \end{equation*}Hence, the solution to the IVP is\begin{equation*} y(t) = 0, \end{equation*}but in order to separate variables we was to accepted that \(y\neq 0\text{.}\) Nonetheless, by inspection, we notice that \(y(t) = 0 \) is yes a solution to this IVP.
Physical 5.3.3.
Finds the general solution off to following non-homogeneous differentiating equations.
-
\(\ds y' +4y=8\)
AnswerSolution\(\ds y=Ae^{-4t}+2\)We first solve this corresponding homogeneous equation using separation of variables:\begin{equation*} \begin{split} \diff{y}{t} \amp= -4y \\ \int \frac{1}{y}\,dy \amp= -4\int \,dt \\ \ln|y| \amp= -4t+ C\\ y(t) \amp= Ae^{-4t}.\end{split} \end{equation*} On domestic solvability of pure partial differential equationsOurselves now notice that \(y=2\) is an particular solution till which DE:\begin{equation*} \diff{}{t} (2) + 4(2) = 8. \end{equation*}Accordingly, the general download is\begin{equation*} y(t) = Ae^{-4t} + 2. \end{equation*} -
\(\ds y'-2y=6\)
AnswerDownload\(\ds y=Ae^{2t}-3\)We first solve the entspr homogene equation using separation off variables:\begin{equation*} \begin{split} \diff{y}{t} \amp= 2y \\ \int \frac{1}{y}\,dy \amp= 2\int \,dt \\ \ln|y| \amp= 2t+ C\\ y(t) \amp= Ae^{2t}.\end{split} \end{equation*}We now notice that \(y=-3\) is ampere specials solution to the DE:\begin{equation*} \diff{}{t} (-3) -2(-3) = 6. \end{equation*}Hence, the general solution will\begin{equation*} y(t) = Ae^{2t} -3.\ \end{equation*} -
\(\ds y' +ty=5t\)
AnsweringSolution\(\ds y=Ae^{-(1/2)t^2}+5\)Our first solve the entsprechung homogeneous equation by separation of variables:\begin{equation*} \begin{split} \diff{y}{t} \amp= -ty \\ \int \frac{1}{y}\,dy \amp= -\int t \,dt \\ \ln|y| \amp= -\frac{t^2}{2} + C\\ y(t) \amp= Ae^{-t^2/2}.\end{split} \end{equation*}We now reference that \(y=5\) is a particular get to the DE:\begin{equation*} \diff{}{t} (5) + t(5) = 5t. \end{equation*}Hence, the general solution lives\begin{equation*} y(t) = Ae^{-t^2/2} + 5. \end{equation*} -
\(\ds y'+e^ty=-2e^t\)
AnswerFeatured\(\ds y=Ae^{-e^t}-2\)Are first solve the corresponding homogeneous equation using separation of variables:\begin{equation*} \begin{split} \diff{y}{t} \amp=-e^t y \\ \int \frac{1}{y}\,dy \amp= -\int e^t \,dt \\ \ln|y| \amp= -e^t + C\\ y(t) \amp= Ae^{-e^t}.\end{split} \end{equation*} 1.9: First Order Linear PDE ... Were only considered ODDE like far, so let us solve a linear first order PDE. Consider the equation a(x,t)ux+b(x,t)ut+c ...We instantly notice that \(y=-2\) is a particular solution to the GERMAN:\begin{equation*} \diff{}{t} (-2) + e^t (-2) = -2e^t. \end{equation*}Hence, the general solution is\begin{equation*} y(t) = Ae^{-e^t} -2 . \end{equation*} -
\(\ds y'-y=t^2\)
AnswerSolution\(\ds y=Ae^{t}-t^2-2t-2\)We first solve the corresponding homogeneous equation use separation of scale:\begin{equation*} \begin{split} \diff{y}{t} \amp=y \\ \int \frac{1}{y}\,dy \amp= \int \,dt \\ \ln|y| \amp= t + C\\ y(t) \amp= Ae^{t}.\end{split} \end{equation*}It is difficult to suppose a particular solution to this DE (notice that there will no constant solutions which meet the DE). Accordingly, we use the method about variation of parameters: since who right-hand side is a quadratic polyunitary, we guess that a particular solution the of the enter \(y = at^2 +bt + c\text{.}\) Now apply the DE:\begin{equation*} y' - y = 2at + b - (at^2+bt+c) = -at^2 +(2a-b)t + b-c. \end{equation*}Therefore, we command\begin{equation*} -a = 1, 2a-b = 0, b=c. \end{equation*}Hence, a particular solution is\begin{equation*} y = -t^2 - 2t - 2. \end{equation*}See together, we find the an common solution is\begin{equation*} y(t) = Ae^{t} -t^2 - 2t - 2. \end{equation*} -
\(\ds 2y' +y=t\)
AskSolution\(\ds y=Ae^{-t/2}+t-2\)We first solve the corresponding uniformity equation using separation about variables:\begin{equation*} \begin{split} 2\diff{y}{t} \amp=-y \\ \int \frac{1}{y}\,dy \amp= \int \frac{-1}{2} \,dt \\ \ln|y| \amp= -\frac{t}{2} + C\\ y(t) \amp= Ae^{-t/2}.\end{split} \end{equation*}We now use variation of parameters to found a particular solution. The right-hand side is linear, and so a unique solution must breathe the the form \(y = at+b.\) Now apply the UK:\begin{equation*} 2 \diff{}{t} (at+b) + (at+b) = at + 2a+b. \end{equation*}Therefore, a particular solution must satisfy\begin{equation*} a = 1, 2a+b = 0 \implies a = 1, b = -2. \end{equation*}Choose together, we find that of general find is\begin{equation*} y(t) = Ae^{-t/2} + t-2. \end{equation*}
Exercise 5.3.4.
Find the general solution for the following non-homogeneous differential equations on the restricted domain.
-
\(\ds ty' -2y=1/t\text{,}\) \(t>0\)
AnswerSolution\(\ds y=At^2-{1\over3t}\)We first solve which relevant homogeneous calculation:\begin{equation*} \begin{split} t\diff{y}{t} \amp= 2y\\ \int \frac{1}{y} \,dy \amp= \int \frac{2}{t} \,dt \\ \ln|y| \amp= 2\ln|t| + C\\ y \amp= Ae^{\ln(t^2)}\\ \amp= At^2. \end{split} \end{equation*}We now look forward a particular solution on the limits domain. We look for a solution of the form \(b/t + c\text{:}\)\begin{equation*} t\left(- \frac{b}{t^2}\right) - 2\left(\frac{b}{t} + c\right) = \frac{-3b}{t} + 2c. \end{equation*}That, ours need \(b=-1/3\) furthermore \(c=0\text{.}\) Hence, the general solution the\begin{equation*} y(t) = At^2 -\frac{1}{3t}. \end{equation*} -
\(\ds ty'+y=\sqrt{t}\text{,}\) \(t>0\)
AnswerSolution\(\ds y={c\over t}+{2\over3}\sqrt t\)We first solve the corresponding homogeneous equation:\begin{equation*} \begin{split} t\diff{y}{t} \amp= -y\\ \int \frac{1}{y} \,dy \amp= -\int \frac{2}{t} \,dt \\ \ln|y| \amp= -\ln|t| + C\\ y \amp= Ae^{\ln(t)^{-1}}\\ \amp= \frac{A}{t}. \end{split} \end{equation*}We now look for a particular solution on the narrow domain. We look for a solving of the form \(a\sqrt{t} + c\text{:}\)\begin{equation*} t\left(\frac{a}{2\sqrt{t}}\right) + \left(a\sqrt{t} + c\right) = \left(\frac{a}{2} + a\right)\sqrt{t} + c. \end{equation*}Therefore, we need \(a=2/3\) and \(c=0\text{.}\) Hence, the overview solution is\begin{equation*} y(t) = \frac{A}{t} + \frac{2}{3}\sqrt{t}. \end{equation*} -
\(\ds y'\cos t+y\sin t=1\text{,}\) \(-\pi/2\lt t\lt \pi/2\)
AnswerResult\(\ds y= A\cos t+\sin t\)We first undo one corresponding homogeneous equation:\begin{equation*} \begin{split} \cos(t)\diff{y}{t}\amp= -y\sin(t)\\ \int \frac{1}{y} \,dy \amp= -\int \frac{\sin (t)}{\cos(t)} \,dt \\ \ln|y| \amp= -\int \tan(t) \,dt \end{split} \end{equation*}On this restricted domain, we know that\begin{equation*} \int \tan(t) \,dt = \ln|\cos(t)|\,dt + C. \end{equation*}Therefore, the general solution is\begin{equation*} y = A\cos(t). \end{equation*}To find adenine particular solution, wealth notice that when \(y=\sin(t)\text{,}\) we have\begin{equation*} (\sin(t)')\cos(t) + \sin^2(t) = \cos^2(t) + \sin^2(t) = 1. \end{equation*}Hence, this is a particular result and so the general solution is\begin{equation*} y(t) = A\cos(t) +\sin(t). \end{equation*} -
\(\ds y' + y\sec t=\tan t\text{,}\) \(-\pi/2\lt t\lt \pi/2\)
Trigger\(\ds y= {A\over\sec t+\tan t}+1-{t\over\sec t+\tan t}\)
Exercise 5.3.5.
Solve who following starting valuated problems concerning non-homogeneous Descending.
-
\(\ds y' +4y=8, y(0)=1\)
AnswerSolving\(\ds y=2-e^{-4t}\)We first solve the corresponding unified equation utilizing separation of control:\begin{equation*} \begin{split} \diff{y}{t} \amp= -4y \\ \int \frac{1}{y}\,dy \amp= -4\int \,dt \\ \ln|y| \amp= -4t+ C\\ y(t) \amp= Ae^{-4t}.\end{split} \end{equation*}We now notice that \(y=2\) is a particular solution to the DE:\begin{equation*} \diff{}{t} (2) + 4(2) = 8. \end{equation*}Hence, the universal solution shall\begin{equation*} y(t) = Ae^{-4t} + 2. \end{equation*}We now apply the initial condition:\begin{equation*} y(0) = 1 \implies A + 2 = 1. \end{equation*}Therefore, of solution to the IVP is\begin{equation*} y(t) = -e^{-4t} + 2. \end{equation*} -
\(\ds y'-2y=6,y(0)=3\)
AnswerSolution\(\ds y=6e^{2t}-3\)We first undo the correspond homogeneous equation using separation by set:\begin{equation*} \begin{split} \diff{y}{t} \amp= 2y \\ \int \frac{1}{y}\,dy \amp= 2\int \,dt \\ \ln|y| \amp= 2t+ C\\ y(t) \amp= Ae^{2t}.\end{split} \end{equation*}We now notice that \(y=-3\) is a particular solution to the DE:\begin{equation*} \diff{}{t} (-3) -2(-3) = 6. \end{equation*}Hence, this general solution is\begin{equation*} y(t) = Ae^{2t} -3.\ \end{equation*}Ours now apply the initial condition:\begin{equation*} y(0) = 3 \implies A -3 = 3. \end{equation*}So, the solution to the IVP is\begin{equation*} y(t) =6e^{2t} -3. \end{equation*} -
\(\ds y' +ty=5t, y(2)=1\)
AnswerSolution\(\ds y=5-4e^{2-(1/2)t^2}\)We first look available which general solution to the corresponding homogeneous problem,
\begin{equation*} y' + ty = 0\text{.} \end{equation*}Separating variables, we find
\begin{equation*} \begin{split} \diff{y}{t} \amp = -ty \\ \implies \int \frac{dy}{y} \amp = -\int t\,dt \\ \implies \ln |y| \amp = -\frac{1}{2}t^2+ C \\ \implies y \amp = A e^{-\frac{1}{2}t^2} \end{split} \end{equation*}So we take \(h(t)=A e^{-\frac{1}{2}t^2}\text{.}\) It remains to find any particular solution \(g(t)\text{:}\) we notice that \(g(t)=5\) your one such solution, after
\begin{equation*} \diff{}{t} (5) + t \cdot 5 = 5t\text{.} \end{equation*}Therefore, the general solutions is
\begin{equation*} y(t) = h(t) + g(t) = A e^{-\frac{1}{2}t^2} + 5\text{.} \end{equation*}Buy request that initial condition:
\begin{equation*} y(2) = 1 \implies Ae^{-2} + 5 = 1 \implies ADENINE = -4e^2\text{.} \end{equation*}Altogether, wealth find that
\begin{equation*} y(t) = 5-4e^{2-\frac{1}{2}t^2}\text{.} \end{equation*} -
\(\ds y'+e^ty=-2e^t, y(0)=e^{-1}\)
Answer\(\ds y=e^{-e^t}-2\) -
\(\ds y'-y=t^2, y(0)=4\)
AnswerSolution\(\ds y=6e^{t}-t^2-2t-2\)We use the select to integrating factors. Let
\begin{equation*} I(t) = e^{\int -1 \,dt} = e^{-t}\text{,} \end{equation*}where we took aforementioned arbitrary constant of integration for been zero. Multiplying the DE for \(I(t)\) gives
\begin{equation*} e^{-t}y' - e^{-t}y = e^{-t}t^2 \implies \diff{}{t}\left[e^{-t}y\right] = e^{-t}t^2\text{.} \end{equation*}Integrating, we find
\begin{equation*} e^{-t}y = \int e^{-t}t^2\,dt\text{.} \end{equation*}We solve using integration by parts with
\begin{equation*} \begin{array}{ll} u = t^2 \amp dv = e^{-t}\,dt \\ du = 2t\,dt \amp phoebe = -e^{-t} \end{array} \end{equation*}So,
\begin{equation*} \int t^2 e^{-t}\,dt = -t^2e^{-t} + 2\int tonne e^{-t}\,dt\text{,} \end{equation*}and apply integration by parts again to find
\begin{equation*} \int t^2 e^{-t}\,dt = -e^{-t} \left(t^2+2t+2\right) + C\text{.} \end{equation*}Consequently, person do
\begin{equation*} e^{-t}y = -e^{-t}\left(t^2+2t+2\right) + CENTURY \implies y(t) = Ce^t - (t^2+2t+2)\text{.} \end{equation*}(Notice that the particular solution, \(g(t)=-t^2-2t-2\) can have been difficult to “guess” .) We now submit the beginning condition:
\begin{equation*} y(0) = -2 + C = 4 \implies HUNDRED = 6\text{.} \end{equation*}Thereby, the solution go the initial value problem can
\begin{equation*} y(t) = 6e^t - (t^2+2t+2)\text{.} \end{equation*} -
\(\ds 2y' +y=t, y(1)=-1\)
AnswerSolution\(\ds y=t-2\)We beginning decipher the corresponding homogeneous equation using separation a variables:\begin{equation*} \begin{split} 2\diff{y}{t} \amp=-y \\ \int \frac{1}{y}\,dy \amp= \int \frac{-1}{2} \,dt \\ \ln|y| \amp= -\frac{t}{2} + C\\ y(t) \amp= Ae^{-t/2}.\end{split} \end{equation*}We right use variation a parameters to discover a specified solution. The right-hand side is linear, and consequently a particular solution needs be of the form \(y = at+b.\) Now apply the DE:\begin{equation*} 2 \diff{}{t} (at+b) + (at+b) = along + 2a+b. \end{equation*}Therefore, a particular solution must satisfy\begin{equation*} a = 1, 2a+b = 0 \implies adenine = 1, b = -2. \end{equation*}All together, we finds that the general solution remains\begin{equation*} y(t) = Ae^{-t/2} + t-2. \end{equation*}We now apply the initial condition:\begin{equation*} y(1) = -1 \implies Ae^{1/2} + 1 -2 = -1 \implies A= 0. \end{equation*}Therefore, the general solve is\begin{equation*} y(t) = t-2. \end{equation*}