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Assume $x(t)$ be the solution to the differential equation $x''+a(t)x'+b(t)x=0$ where $a,b$ are continuous functions.

I need to show that $y$ is a second and linear independent solution of the equation such that $y(t)=x(t)u(t)$ and $u(t)$ is a non constant solution to $u''+(2(x'/x)+a)u'=0$.

So far I have used the substitution that $v=u'$ and create a first order linear equation to solve. However, what method is most appropriate to solve $u'$? I was thinking separation by variables.

Any help thanks :)

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You don't need to solve the equation. Just substitute. Implicit in the above is $x(t) \neq 0$ for all $t$ (otherwise $u$ is ill-defined).

I will suppress '$t$' dependence for sanity. If $y = ux$, then $\dot{y} = \dot{x}u+\dot{u}x$ and $\ddot{y} = \ddot{x}u+2 \dot{x}\dot{u}+\ddot{u}x$. Then expand $\ddot{y}+a \dot{y}+b y$.

$\ddot{y}+a \dot{y}+b y = \ddot{x}u+2 \dot{x}\dot{u}+\ddot{u}x + a\dot{x}u+a\dot{u}x+b x u = u(\ddot{x}+a \dot{x}+ bx)+x(2 \frac{\dot{x}}{x}\dot{u}+\ddot{u}+a \dot{u})$ which is zero since the terms in parentheses are zero. This shows that $y$ is a solution.

To show that $x,y$ are linearly independent, suppose $\alpha x + \beta y = 0$ (ie, identically zero) for some constants $\alpha, \beta$. Substituting gives $x(\alpha+\beta u) = 0$. Now argue that $\alpha=\beta = 0$.

Since $x(t) \neq 0$ for all $t$ we have $\alpha+\beta u = 0$, which implies that $u$ is constant, which contradicts the assumptions. Hence $\alpha=\beta = 0$, and $x,y$ are linearly independent.

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