Solution to the ODE $\qquad2\frac{\mathrm dI(a)}{\mathrm da}+aI(a)=0$

Question

I am working on this question

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Show that the function $$I(a)=\int\limits_0^\infty e^{-u^2}\cos(au)\,\mathrm du $$ satisfies the differential equation $$2\frac{\mathrm dI(a)}{\mathrm da}+aI(a)=0$$

Hence find an expression for $I(a)$

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This is my working so far

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$$\begin{align*} \frac{\mathrm dI(a)}{\mathrm da}&=\int\limits_0^\infty \frac{\partial}{\partial a}e^{-u^2}\cos(au)\,\mathrm du\\ &=\int\limits_0^\infty -ue^{-u^2}\sin(au)\,\mathrm du\\ \\ \end{align*}$$

$$\begin{align*} 2\frac{\mathrm dI(a)}{\mathrm da}+aI(a) &=\int\limits_0^\infty -2ue^{-u^2}\sin(au)\,\mathrm du+\int\limits_0^\infty ae^{-u^2}\cos(au)\,\mathrm du\\ &=\int\limits_0^\infty\frac{\mathrm d }{\mathrm d u}\left(e^{-u^2}\sin(au)\right)\mathrm du\\ &=\left.e^{-u^2}\sin(au)\right|_0^\infty\\ &=0 \end{align*} $$

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$$\begin{align*} I(0)&=\int\limits_0^\infty e^{-u^2}\,\mathrm du\\ \text{let }u=v^{1/2}\\ \mathrm du=\frac{v^{-1/2}}2\,\mathrm dv\\ &=\int\limits_0^\infty e^{-v}\frac{v^{-1/2}\,\mathrm dv}2\\ &=\frac12\int\limits_0^\infty {v^{-1/2}e^{-v}\,\mathrm dv}\\ &=\frac{\Gamma(\tfrac12)}2\\ &=\frac{\sqrt\pi}2 \end{align*}$$

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I'm not sure what the question means by

"find an expression for $I(a)$"

How should I proceed?

Answer 1

You can check here that $\Gamma(1/2) = \sqrt{\pi}$. So you know that $$2 I'(a)+aI(a)=0, \qquad I(0) =\sqrt{\pi}.$$

You can solve this differential equation via separation of variable, i.e., rewrite the equation as $$\log'I(a)=\frac{I'(a)}{I(a)} =- \frac{a}{2}.$$ Then you know that $\log I(a) = C - \frac{a^2}4$. With the value at $a=0$ you can determine $C$ and obtain the final solution $$I(a)= \frac{\sqrt{\pi}}2 e^{-a^2/4} .$$

Answer 2

So, to

"find an expression for I(a)"

I just have to use separation of variables on $$2\frac{\mathrm dI(a)}{\mathrm da}+aI(a)=0$$

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\begin{align*} 2\frac{\mathrm dI(a)}{\mathrm da}+aI(a)&=0\\ 2\frac{\mathrm dI(a)}{I(a)}+a\,\mathrm da&=0\\ 2\int\frac{\mathrm dI(a)}{I(a)}+\int a\,\mathrm da&=0\\ 2\ln I(a)+\frac{a^2}2&=c\\ \\ 2\ln I(0)&=c\\ 2\ln\frac{\sqrt\pi}2&=c\\ \ln\frac{\pi}4&=c \\ \\ 2\ln I(a)+\frac{a^2}2&=\ln\frac{\pi}4\\ \ln I(a)&=\ln\sqrt[2]{\frac{\pi}4}-\frac{a^2}4\\ I(a)&=\frac{\sqrt\pi}2e^{-\frac{a^2}4} \end{align*} Which is the solution to the ODE.