UnkleRhaukus
  • UnkleRhaukus
Show that the function satisfies the DE
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\mathrm du\\ \frac{\mathrm dI(a)}{\mathrm da}&=\int\limits_0^\infty \frac{\partial}{\partial a}e^{-u^2}\cos(au)\mathrm du\\ &=-a\int\limits_0^\infty e^{-u^2}\sin(au)\mathrm du\\ \\ \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
now what should i do ?

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BAdhi
  • BAdhi
firstly $$\frac{dI}{da}=\int\limits_{0}^{\infty}-ue^{-u^2}\sin(au)du$$ when we consider the left side of the differential equation, $$\begin{align*}2\frac{dI}{da}+aI &= \int\limits_{0}^{\infty}(-2u)e^{-u^2}\sin(au)du+\int \limits_{0}^{\infty}e^{-u^2}a\cos(au)du\\ &=\int\limits_0^\infty\frac{de^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\sin(au)}{du}du\\ &=\int \limits_0^\infty \frac{d\,e^{-u^2}\sin(au)}{du}\,du\\ &=\int\limits_0^\infty d\,e^{-u^2}\sin(au)\\ &=\left[e^{-u^2}\sin(au)\right]_0^{\infty}=0-0=0 \end{align*}$$
UnkleRhaukus
  • UnkleRhaukus
some kinda reverse integration by parts?
UnkleRhaukus
  • UnkleRhaukus
im not quite sure how you got your third line
UnkleRhaukus
  • UnkleRhaukus
can you explain how you got from \[\begin{align*} \int\limits_{0}^{\infty}(-2u)e^{-u^2}\sin(au)du+\int \limits_{0}^{\infty}e^{-u^2}a\cos(au)du\\ \text {to}\\ =\int\limits_0^\infty\frac{de^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\sin(au)}{du}du\end{align*}\] @BAdhi
BAdhi
  • BAdhi
we know that $$\frac{d\,e^{-u^2}}{du}=(-2u)\,e^{-u^2}$$ and $$\frac{d\,\sin(au)}{du}=a\cos(au)$$ so, $$\begin{align*}\int\limits_0^\infty (-2u)\,e^{-u^2}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}a\cos(au)\,du&=\int\limits_0^\infty \frac{d\,e^{-u^2}}{du}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}\frac{d\,\sin(au)}{du}\,du\\ &=\int\limits_0^\infty \underbrace{\frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}}_{(1)}\,du\end{align*}$$ (1) can be reduced from the differentiation of a multiplication, property (i.e.) $$\frac{d\,uv}{dx}=u\frac{du}{dx}+v\frac{dv}{dx}$$ $$\frac{d\,\left(e^{-u^2}\sin(au)\right)}{du}=\frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}$$ (1) can be replaced with this, $$\begin{align*}\int\limits_0^\infty (-2u)\,e^{-u^2}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}a\cos(au)\,du&=\int\limits_0^\infty \frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}\,du \\ &=\int\limits_0^\infty\frac{d\,\left(e^{-u^2}\sin(au)\right)}{du}du\end{align*}$$ Is it clear enough for you?
UnkleRhaukus
  • UnkleRhaukus
yes
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} 2\frac{\mathrm dI(a)}{\mathrm da}+aI(a) &=\int\limits_0^\infty -2ue^{-u^2}\sin(au)\mathrm du+\int\limits_0^\infty ae^{-u^2}\cos(au)\mathrm du\\ &=\int\limits_0^\infty\frac{\mathrm d }{\mathrm d u}\left(e^{-u^2}\sin(au)\right)\mathrm du\qquad\text{(using the product rule)}\\ &=\left.e^{-u^2}\sin(au)\right|_0^\infty\\ &=0 \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\,\mathrm du\\ I(0)&=\int\limits_0^\infty e^{-u^2}\,\mathrm du\\ \frac{\sqrt \pi}2&=\int\limits_0^\infty e^{-u^2}\,\mathrm du\\ \text{let }u=v^{1/2}\\ \mathrm du=\frac{v^{-1/2}}2\,\mathrm dv\\ \frac{\sqrt \pi}2&=\int\limits_0^\infty e^{-v}\frac{v^{-1/2}\,\mathrm dv}2\\ {\sqrt \pi}&=\int\limits_0^\infty {v^{-1/2}e^{-v}\,\mathrm dv}\\ {\sqrt \pi}&=\Gamma(\tfrac12)\\ \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
im not sure what they mean by an expression for \(I(a)\)
Sepeario
  • Sepeario
http://mathhelpforum.com/calculus/197830-how-show-function-satisfies-differential-equation.html
UnkleRhaukus
  • UnkleRhaukus
@Sepeario yes i have tried differentiating \(I(a)\)
anonymous
  • anonymous
@UnkleRhaukus integrate the expression using integration by parts technique.then subst for I(0).
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\,\mathrm du\\ \text{let }u=v^{1/2}\\ \mathrm du=\frac{v^{-1/2}}2\,\mathrm dv\\ I(a)&=\int\limits_0^\infty e^{-v}\cos(a\sqrt v)\frac{v^{-1/2}\,\mathrm dv}2\\ \end{align*}\] how do i integrate by parts, there are three terms? do i break up the cos into e^... bits?
sirm3d
  • sirm3d
\[2\frac{ dI }{ da }+aI=0\\2 \frac{ dI }{ I }+a\space da=0\\2lnI+\frac{ a^2 }{ 2 }=C\]when \(\displaystyle a=0,\;I=\frac{\sqrt{\pi}}{2}\)\[2\ln \frac{ \sqrt{\pi} }{ 2 }+\frac{ 0^2 }{ 2 }=C\]\[2\ln I+\frac{a^2}{2}=2\ln\frac{\sqrt{\pi}}{2}\]\[\vdots\\\ln \left[\frac{ 4 }{ \pi }I^2(a)\right]=-\frac{ a^2 }{ 2 }\\I^2(a)=\frac{\pi}{4}\exp(-a^2/2)\]
UnkleRhaukus
  • UnkleRhaukus
Thankyou @sirm3d & @BAdhi \[\begin{align*} 2\frac{\mathrm dI(a)}{\mathrm da}+aI(a)&=0\\ 2\frac{\mathrm dI(a)}{I(a)}+a\,\mathrm da&=0\\ 2\int\frac{\mathrm dI(a)}{I(a)}+\int a\,\mathrm da&=0\\ 2\ln |I(a)|+\frac{a^2}2&=c \end{align*}\] \[\begin{align*} 2\ln |I(0)|&=c\\ 2\ln\frac{\sqrt\pi}2&=c\\ \ln\frac{\pi}4&=c\end{align*}\] \[\begin{align*} 2\ln |I(a)|+\frac{a^2}2&=\ln\frac{\pi}4\\ \ln |I(a)|&=\ln{\frac{\sqrt\pi}2}-\frac{a^2}4\\ I(a)&={\frac{\sqrt\pi}2}e^{-\frac{a^2}4} \end{align*} \]

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