# Relationship between the solutions of a 2nd Order ODE

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 $$y_2(x)=x\cdot y_1(x),\forall x\in \mathbb R.$$

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.