Consider the following ODE:

\begin{equation} (\cos x)y''-2(\sin x)y'-(\cos x)y=e^x \end{equation}

The above equation is a second order linear ODE. However, I noticed that it doesn't have constant coefficients, so I cannot "guess" the solution is of the form $e^{\lambda x}$. Im very confused by this, since I have never solved a 2nd order ODE with variable coefficients. The first part of the question says:


Show that the ODE is of the form:

\begin{equation} \frac{d^2}{dx^2}(f(x)y)=e^x \end{equation}

by finding the function f.


Hence, find the general soluion of he differential equation.

For part (a) I have have no idea where to begin, but I tink that If i knew how to complete part (a) I would be able to find the general solution since I hae done things like that many times. Any help (part a especially) would be much appreciated. Thanks in advance.


a) you have to find $f(x)$ such that $\frac{d^2}{dx^2}(f(x)y)=e^x$, but $$e^x=\frac{d^2}{dx^2}(f(x)y)=\frac{d}{dx}(f'(x)y+f(x)y')=f''(x)y+f'(x)y'+f'(x)y'+f(x)y''$$ So from the equation $$(\cos x)y''-2(\sin x)y'-(\cos x)y=e^x=f''(x)y+2f'(x)y'+f(x)y''$$ we obtain that $f(x)=\cos x$.

b) you can before find the soluction for the associate homogeneous equaction: $$(\cos x)y''-2(\sin x)y'-(\cos x)y=0$$ Using what we have seen in the point a) we have $$0=\frac{d^2}{dx^2}(\cos (x)\cdot y)$$ so $\cos (x) \cdot y$ or it's a costant or a polinomial with $deg=1$. So a base of the associate homogeneous equaction it's $c_1\cdot sec(x)+c_2\cdot xsec(x)$ ($sec(x)=\frac{1}{\cos x}$). Now from $$e^x=\frac{d^2}{dx^2}(f(x)y)$$ we can easily obtain a particular soluction like $sec(x)e^x$. Summing the particular solution with the general soluction of the associate homogeneous equaction you find what you want.

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The differential equation is $y''-2(\tan x )y'-y=e^x\sec x$.Calculate the invariant $I=Q-(1/4)P^2-(1/2)dP/dx$ where $P,Q,R$ have usual meaning and $I=0$ by calculation.So the substitution $y=ve^{-1/2\int Pdx}=v\sec x$ transforms the differential equation into $v''+Iv=Re^{1/2\int P dx}$,i.e. $v''=e^x$,integrating $v=e^x+Ax+B$ and $y=v\sec x$,which is the solution of the second order linear differential equation and here $f(x)=\cos x $ as per your question as $y=v\sec x$ implies $y\cos x=v$ so that $D^2(y\cos x)=e^x$.

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