Reduction of Order via Substitution

Question

Suppose $u_1=\sin{x^2}$ is a solution of $$xu''-u'+4x^3u=0\Rightarrow u''-x^{-1}u'+4x^2u=0 \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ I am trying to find a second, linearly independent, solution (say $u_2$) to the above equation using reduction of order.

Now, using the formula $$u_2=u_1\int\frac{dx}{u^2_1\text{exp}\left(\int p(x)\ dx\right)} \ \ \ \ ,\ p(x)=-x^{-1}$$ I have found that $$u_2=\frac{\sin{x^2}}{2}\int \text{cosec}^2x \ dx=-\frac{\cos{x^2}}{2}$$ Now I tried to replicate this using the substitution $u_2=u_1v(x)$. After finding the first and second derivatives of this equation and substituting into $(1)$, I get $$v''\sin{x^2}+v'\left(4x\cos{x^2}-x^{-1}\sin{x^2}\right)=0$$ Letting $w=v'$, $$\frac{dw}{dx}\sin{x^2}+w\left(4x\cos{x^2}-x^{-1}\sin{x^2}\right)=0$$ I tried to simplify this using the integrating factor $$\text{exp}\left(\int 4x\cos{x^2}-x^{-1}\sin{x^2} \ dx\right)$$ but could not compute it.

How can I solve this problem using my suggested substitution?

Answer 1

The ODE for $w$ is actually separable: \begin{align*} \sin(x^2)\frac{dw}{dx} + \left(4x\cos(x^2) - \frac{\sin(x^2)}{x}\right)w & = 0 \\ \int \frac{1}{w}\, dw & = \int \frac{1}{\sin(x^2)}\left(\frac{\sin(x^2)}{x} - 4x\cos(x^2)\right)\, dx \\ \int\frac{1}{w}\, dw & = \int\frac{1}{x} - 4x\cot(x^2)\, dx \\ \end{align*} Making a change of variable $s = x^2$ on the second integrand, we obtain \begin{align*} \int\frac{1}{w}\, dw & = \int \frac{1}{x}\, dx - 2\int\cot(s)\, ds \\ \ln|w| & = \ln|x| - 2\ln|\sin(x^2)| + C \end{align*} You may then exponentiate each side and get $$ |w| = e^C\frac{|x|}{\sin^2(x^2)} = \frac{A|x|}{\sin^2(x^2)}. $$

Answer 2

Here is a solution without the reduction method. Define the operators $D$ and $X$ by $(D\,h)(x):=h'(x)$ and $(X\,h)(x):=x\,h(x)$. For any function $\phi$, we also define the operator $\phi(X)$ to be $\big(\phi(X)\,h\big)(x):=\phi(x)\,h(x)$. Observe that $$D^2-\frac1X\,D+4X^2=\left(D+2\text{i}\,X-\frac{1}{X}\right)\,(D-2\text{i}\,X)\,,$$ where $\text{i}:=\sqrt{-1}$.

If $u$ is a solution to $\displaystyle \left(D^2-\frac1X\,D+4X^2\right)\,u=0$, we set $v:=(D-2\text{i}\,X)\,u$, so that $$\left(D+2\text{i}\,X-\frac{1}{X}\right)\,v=0\,,\text{ or }D\,\left(\frac{\exp(\text{i}\,X^2)}{X}\,v\right)=0\,.$$ Thus, $v(x)=-4\text{i}\,a\,x\,\exp\left(-\text{i}x^2\right)$ for some constant $a$. Now, $(D-2\text{i}\,X)\,u=v$ gives $$D\,\big(\exp(-\text{i}\,X^2)\,u\big)=\exp(-\text{i}\,X^2)\,v\,,\text{ whence }u(x)=\exp(+\text{i}\,x^2)\,\int\,\exp(-\text{i}\,x^2)\,v(x)\,\text{d}x\,.$$ Consequently, for some constant $b$, we get $$u(x)=\exp(+\text{i}\,x^2)\,\int\,(-4\text{i}\,a\,x)\,\exp\left(-2\text{i}\,x^2\right)\,\text{d}x=\exp(+\text{i}\,x^2)\,\big(a\,\exp(-2\text{i}\,x^2)+b\big)\,.$$ That is, $$u(x)=a\,\exp(-\text{i}\,x^2)+b\,\exp(+\text{i}\,x^2)=(a+b)\,\cos(x^2)-\text{i}\,(a-b)\,\sin(x^2)\,.$$