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Let $y_1,y_2$ form a foundamental set of solutions of the 2nd order linear and homogenous ODE: $$ \ddot y+ \cos(x)\dot y+\frac{\cos^2(x)-2\sin(x)}{4}y=0\quad(*) $$ Show that $y_2(x)=x\cdot y_1(x),\forall x\in \mathbb R.$

My attempt goes like this: Let $y_1\neq0$ be a solution of $(*)$ meaning that $$ \ddot y_1+\cos(x)\dot y_1+\frac{\cos^2(x)-2\sin(x)}{4}y_1=0 $$ and $u(x)=x\cdot y_1(x) ,\forall x\in \mathbb R.$ Then $$ \ddot u+\cos(x)\dot u+\frac{\cos^2(x)-2\sin(x)}{4}u=x\{\ddot y_1+\cos(x)\dot y_1+\frac{\cos^2(x)-2\sin(x)}{4}y_1\}+2\dot y_1+\cos(x)y_1=2\dot y_1+\cos(x)y_1.$$

I am stacked here at this point. I have tried using Abel's Type for the Wronskian of $y_1,y_2$: $W[y_1,y_2](x)=Ce^{-\int \cos(x)dx}=Ce^{-\sin(x)}$ but I cannot show that $2\dot y_1+\cos(x)y_1=0$.

Any suggestions?

A correction made driven by the observation of Gyu Eun Lee

Show that there exist solutions $y_1,y_2$ such that $y_2(x)=x\cdot y_1(x),\forall x\in \mathbb R.$

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I suspect that you are missing an assumption, as the claim is false. If $y_1$ and $y_2$ form a fundamental set of solutions, then so does $y_1$ and $y_2 + Cy_1$ for any constant $C$. So it is impossible to conclude that $y_2 = xy_1$ from the conclusions given here: $y_2 = (x+1)y_1$ would be an equally valid conclusion.

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  • $\begingroup$ You are right.I reformed the question properly $\endgroup$ Commented Apr 30, 2021 at 5:31

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