Find the solution to the first order ODE's

Question

Find the solution to

$x′= y−x+t $ and $y'= y $

if $x(0)=8$ and $y(0)=2$

I'm confused how to go about this question. If anyone could offer some clarification that would be greatly appreciated!

Answer 1

HINT

The system is in the form

$$ \begin{bmatrix} x^\prime \\y^\prime \end{bmatrix} = \begin{bmatrix} -1 & 1\\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} +t\begin{bmatrix} 1 \\0 \end{bmatrix} \quad \vec x^{\prime}=A \vec x+t\vec b$$

Firstly we need to find the complementary solution to $\vec x^{\prime}=A \vec x$ assuming $\vec x=\vec u\cdot e^{at}$ that is

$$a\cdot \vec u\cdot e^{at}=A\vec u\cdot e^{at}\implies(A-aI)\vec u=0$$

and from here we can find the eigenvalues, the eigenvectors and the complementary solution $\vec x_c$ to the system.

Then we need to find the particular solution to $\vec x^{\prime}=A \vec x+t\vec b$ assuming

$$\vec x_p=t\vec v + \vec w \implies \vec x_p^{\prime}=\vec v$$

and then

$$\vec v=tA\vec v+A\vec w+t\vec b\implies t(A\vec v+\vec b)+(A\vec w-\vec u)=0 $$

that is

• $A\vec v+\vec b=\vec 0$ from which we find $\vec v$
• $A\vec w-\vec v=\vec 0$ from which we find $\vec w$

Then the general solution is $\vec x=\vec x_c+\vec x_p$ and we can impose the initial condition to find the constants in $\vec x_c$.

Here you can find some worked example.

Answer 2

First solve the second equation as it does not contain $x$. Then insert $$ y(t)=2e^t $$ into the first equation and solve via your preferred method, integrating factor, homogeneous plus particular solution etc.

Answer 3

First solve the second equation $$y'=y \implies \ln |y|=t+K_1 \implies y=K_1e^t$$ Then solve first equation $$x′= y−x+t$$ $$x'+x=K_1e^t+t$$ Multiply by $e^t$ $$(xe^t)'=K_1e^{2t}+te^t$$ Integrate $$xe^t= \int K_1e^{2t}+te^t dt $$ $$......$$