Solving system of ODE with initial value problem (IVP)

Question

I have a question about a system of ODE. If we have:

$\frac{dx}{dt}=x+2y$

$\frac{dy}{dt}=3x+2y$

with $x(0)=6$ and $y(0)=4$,

how come the solution to the IVP is:

$x(t)=4e^{4t}+2e^{-t}$

$y(t)=6e^{4t}-2e^{-t}$

I tried doing integral by separating the variable but I didn't get that solution. That example & solution are from my numerical method book, please see the attached image

example & solution

Answer 1

take laplace transform on both sides of both differential equations to get

$(s-1)X(s)=2Y(s)+6 \ \ $ ;

$(s-2)Y(s)=3X(s)+4\\$ respectively

solve for $X(s)$ and $Y(s)$ like algebric equations to give ;

$Y(s)=\dfrac{4s+14}{(s+1)(s-4)}\implies y(t)=-2e^{-t}+6e^{4t}$

$X(s)=\dfrac{6s-4}{(s+1)(s-4)}\implies x(t)=2e^{-t}+4e^{4t}$

Answer 2

Hint:

One possibility is to eliminate one of the unknowns.

$$x'=x+2y\implies x''=x'+2y'$$ and $$y'=3x+2y=3x+(x'-x),$$

hence

$$x''-3x'-4x=0.$$

Answer 3

Hint: With $$y=\frac{x'-x}{2}$$ we get $$y'=3x+\frac{3}{2}(x'-x)$$ and we get $$y'=\frac{x''-x'}{2}$$ so we obtain $$\frac{x''-x'}{2}=3x+\frac{3}{2}(x'-x)$$ Can you finish? From here you will get $$x''-4x'-3x=0$$ make the ansatz $$x=e^{\lambda t}$$

Answer 4

Let us consider your system of ODE: $$\frac{dx}{dt}=x+2y\\\frac{dy}{dt}=3x+2y$$ It follows that $$\frac{dx}{dt}+\frac{dy}{dt}=4x+4y\rightarrow \frac{d(x+y)}{dt}=4(x+y)\rightarrow \frac{d(x+y)}{x+y}=4dt$$ Integration on both sides yields $$\ln(x+y)=4t+C\rightarrow x+y=Ce^{4t}\rightarrow y=Ce^{4t}-x$$ Substitution into $\frac{dx}{dt}$ gives $$\frac{dx}{dt}=x+2(Ce^{4t}-x)\rightarrow \frac{dx}{dt}+x=2Ce^{4t}$$ Multiply by $e^t$ on both sides: $$\frac{dx}{dt}e^{t}+xe^{t}=2Ce^{5t}\rightarrow\frac{d(xe^{t})}{dt}=2Ce^{5t}$$ Integration on both sides yields $$xe^{t}=\frac{2C}{5}e^{5t}+K\rightarrow x=\frac{2C}{5}e^{4t}+Ke^{-t}$$ Substitution into $y$ gives $$y=\frac{3C}{5}e^{4t}-Ke^{-t}$$ Apply conditions: $$x(0)+y(0)=6+4=10\rightarrow Ce^{4(0)}=10\rightarrow C=10\\x(0)=6\rightarrow \frac{2(10)}{5}e^{4(0)}+Ke^{-(0)}=6\rightarrow K=2$$ Thus, $$x(t)=4e^{4t}+2e^{-t}\\y(t)=6e^{4t}-2e^{-t}$$

Answer 5

Observe that you have $$ \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \cdot \begin{pmatrix}x \\ y \end{pmatrix} \text{.} $$ The eigenvalues of this matrix are $4, -1$, so both solutions are of the form $a \mathrm{e}^{4x} + b \mathrm{e}^{-1x}$. As others have shown, you then match the coefficients to the initial value data.