Finding complementary function of non-constant coefficients ODE

Question

Determine the complementary function for the homogeneous equation which satisfies the given initial conditions.

$(3+t)x''(t)+(2+t)x'(t)-x(t)=(3+t)^2 \space\space \space \forall \space t > 0 $ subject to $x(0)=0$ and $x'(0)=1$

One independent solution to the homogeneous equation is $e^{-t}$

How would I find the CF of a non-constant coefficients ODE?

Answer 1

You can solve the equation by the method of reduction of order since you already have a solution of the equation. $$x(t)=e^{-t}v(t)$$ I got for the homogeneous equation the following solution $$x(t)=c_1e^{-t}+c_2(t+2)$$ With the method of reduction of order you should be able to solve the inhomogeneous equation..

After substitution I got a first order DE $$(t+3)v''-(t+4)v'=e^t(t+3)^2$$ $$\frac {(t+3)v''-v'}{(t+3)^2}-\frac {v'}{t+3}=e^t$$ $$\left (\frac {v'}{(t+3)}\right )'-\frac {v'}{t+3}=e^t$$ $$\left (\frac {v'}{(t+3)}\right )'e^{-t}-\frac {v'}{t+3}e^{-t}=1$$ $$\left (\frac {v'}{(t+3)} e^{-t}\right)'=1$$ Integrate $$\left (\frac {v'}{(t+3)} e^{-t}\right)=t+K$$ $$ v'=e^{t}(t+K)(t+3)$$ Integrate to get $v(t)$ $$ \implies v(t)=\int e^{t}(t+K)(t+3)dt$$

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Edit after @Lutzl 's right comment

$$(3+t)x''(t)+(2+t)x'(t)-x(t)=(3+t)^2$$ $$(3+t)x''(t)+(2+t)x'(t)+\color{red}{x'-x'}-x(t)=(3+t)^2$$ $$(3+t)(x''+x'(t))-(x'+x(t))=(3+t)^2 $$ Substitute $v=x'+x$ the equation is a first order one. $$(3+t)v'(t)-v(t)=(3+t)^2$$ It's of the form $(\frac f g)'=1$ $$\implies \left (\frac v {t+3}\right)'=1$$ After integration $$v=(t+3)(t+K) \implies x'+x= (t+3)(t+K)$$ $$(xe^t)'=(t+3)(t+K)e^t$$ Integrate $$xe^t=\int (t+3)(t+K)e^t dt$$ Finally $$\boxed {x(t)=Ce^{-t}+K(t+2)+t^2-3}$$

Answer 2

The order reduction approach tells you that if you are given a solution $y_1(t)$, then you search for the second solution in the form $y_2(t)=y_1(t)u(t)$. Here that gives $$ (3+t)u''-(4+t)u'=0\implies u'=Ce^t(3+t),~~u=Ce^t(2+t)+D. $$ This gives the second basis function as $y_2(t)=(2+t)$.

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Or take the first basis solution as inspiration to set $v=y'+y$ to get the reduced equation $$ \frac{d}{dt}\left(\frac{v(t)}{3+t}\right)=1\implies v=(3+t)(C+t),\\~~y=De^{-t}+e^{-t}\int(3+t)(C+t)e^t\,dt=C(2+t)+De^{-t}+e^{-t}\int(3+t)te^t\,dt $$