Using Wronskian to solve nonhomegeneous ODE

Question

I have the given ODE:

$$y''+2y'+2y=e^{-x}\sin x$$

This has the homogeneous solution $y_h=C_1\cos(i-1)x+C_2\sin(-i-1)x$.

The particular solution, in the form $y_p=uy_1+vy_2$, we seek the Ansatz: $y_p=uy_1+vy_2=e^{-x}(\sin x+\cos x)$. So $y_1=e^{-x}\sin x$ and $y_2=e^{-x}\cos x$

Then we aim to solve for $u$ and $v$ by use of the variation of parameters formula:

$$u'y_1+v'y_2=0$$ $$u'y_1'+v'y_2'=f(x)$$

where $f(x)=e^{-x}\sin x$.

So here I should use the Wronskian to facilitate the process. The Wronskian is naturally dependent on $y_1$ and $y_2$ and are $y_1=e^{-x}\sin x$ and $y_2=e^{-x}\cos x$.

So the Wronskian would be

\begin{equation} \text{Det}\begin{vmatrix} e^{-x}\sin x & e^{-x}\cos x\\ e^{-x}\cos x-e^{-x}\sin x & -e^{-x}\cos x-e^{-x}\sin x \end{vmatrix} \end{equation}

My calculation gives:

$Det=-e^{-2x}\cos2x$

So how is this useful to solve the ODE, when I could just use the formula for variation of parameters?

Thanks

Answer 1

$$y''+2y'+2y=0$$ The solution of the homogeneous should be: $$(r+1)^2-i^2=(r+1-i)(r+1+i)=0$$ $$y_h=e^{-x}(c_1\cos x+c_2\sin x)$$ Note that it's easier to first rewrite the DE as: $$y''+2y'+2y=e^{-x}\sin x$$ $$(ye^x)''+ye^x=\sin x$$ $$v''+v=\sin x$$ Then apply the method of variation of parameters.

Answer 2

It is generally easier to solve linear ODE with constant coefficients this way:

• your homogeneous equation has root $r$ with multiplicity $m$ .
• the full equation has a RHS of the form $P(x)e^{rx}$ with $P$ polynomial.

Then you need to search for a particular solution in the form $Q(x)e^{rx}$ with $Q$ polynomial and $$\deg(Q)=\deg(P)+m$$

Although since the homogeneous solution will already have vanishing terms $(C_0+C_1x+\cdots+C_{m-1}x^{m-1})e^{rx}$, you can ignore them in the polynomial Q.

So here your homogeneous equation is $y''+2y'+2y=0$ of characteristic equation $$r^2+2r+2=0$$

It has roots $r=-1\pm i$ of multiplicity $m=1$.

The sinus can be rewritten as a combination of $e^{ix}$ and $e^{-ix}$, therefore you RHS is $$\underbrace{\frac 1{2i}}_{P_1}e^{-x+ix}-\underbrace{\frac 1{2i}}_{P_2}e^{-x-ix}$$

Both terms of your RHS collide with the roots of the characteristic equation so since $P_1,\ P_2$ are constant polynomials, you need to search for particular solutions of the form $(ax+b)e^{-x+ix}$ and $(cx+d)e^{-x-ix}$

(i.e. $\deg(Q)=\deg(P)+m=0+1=1$)

As said coefficients $b$ and $d$ can be ignored because we know already reporting in the ODE, the associated part will just vanish.

Therefore we can search for $ax\,e^{-x+ix}+cx\,e^{-x-ix}$ or directly in recomposed trig form $$\big(A\cos(x)+B\sin(x)\big)xe^{-x}$$

Solving gives $A=-\frac 12$ and $B=0$.