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

$fracdxdt=x+2y$

$fracdydt=3x+2y$

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

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 ;

$Y(s)=dfrac4s+14(s+1)(s-4)implies y(t)=-2e^-t+6e^4t$

$X(s)=dfrac6s-4(s+1)(s-4)implies x(t)=2e^-t+4e^4t$

Hint: With $$y=fracx'-x2$$ we get
$$y'=3x+frac32(x'-x)$$ and we get $$y'=fracx''-x'2$$ so we obtain
$$fracx''-x'2=3x+frac32(x'-x)$$
Can you finish?
From here you will get
$$x''-4x'-3x=0$$ make the ansatz $$x=e^lambda t$$

Hint:

One possibility is to eliminate one of the unknowns.

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

hence

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

Let us consider your system of ODE:
$$fracdxdt=x+2y\fracdydt=3x+2y$$
It follows that
$$fracdxdt+fracdydt=4x+4yrightarrow fracd(x+y)dt=4(x+y)rightarrow fracd(x+y)x+y=4dt$$
Integration on both sides yields
$$ln(x+y)=4t+Crightarrow x+y=Ce^4trightarrow y=Ce^4t-x$$
Substitution into $fracdxdt$ gives
$$fracdxdt=x+2(Ce^4t-x)rightarrow fracdxdt+x=2Ce^4t$$
Multiply by $e^t$ on both sides:
$$fracdxdte^t+xe^t=2Ce^5trightarrowfracd(xe^t)dt=2Ce^5t$$
Integration on both sides yields
$$xe^t=frac2C5e^5t+Krightarrow x=frac2C5e^4t+Ke^-t$$
Substitution into $y$ gives
$$y=frac3C5e^4t-Ke^-t$$
Apply conditions:
$$x(0)+y(0)=6+4=10rightarrow Ce^4(0)=10rightarrow C=10\x(0)=6rightarrow frac2(10)5e^4(0)+Ke^-(0)=6rightarrow K=2$$
Thus,
$$x(t)=4e^4t+2e^-t\y(t)=6e^4t-2e^-t$$

Observe that you have
$$ beginpmatrix dotx \ doty endpmatrix = beginpmatrix 1 & 2 \ 3 & 2 endpmatrix cdot beginpmatrixx \ y endpmatrix text. $$
The eigenvalues of this matrix are $4, -1$, so both solutions are of the form $a mathrme^4x + b mathrme^-1x$. As others have shown, you then match the coefficients to the initial value data.