Question

Prove Euler's Equation
\[x^2\frac{\text d^2y}{\text d{x^2}}+\alpha x\frac{\text dy}{\text dx}+by=f(x)\]\[\Downarrow\qquad\Downarrow\]\[\frac{\text d^2y}{\text d{t^2}}+(\alpha-1)\frac {\text dy}{\text dt}+by=g(t)\]

Answer 1

let \(\ln x=t\)
\[x=e^t\]\[\frac{\text dx}{\text dt}=e^t\qquad \frac{\text d^2x}{\text dt^2}=e^t\]

Answer 2

\[\frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}=e^{-t} \frac{dy}{dt}\]
\[\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dt}(\frac{dy}{dx})\frac{dt}{dx}=e^{-t} \frac{d}{dt}(e^{-t} \frac{dy}{dt})=e^{-2t}(\frac{d^2y}{dt^2})-e^{-t} \frac{dy}{dt}\]

Answer 3

//

Answer 4

plug in the original equation

\[e^{2t} (e^{-2t} \frac{d^2y}{dt^2}-e^{-2t}\frac{dy}{dt})+\alpha e^{t} (e^{-t} \frac{dy}{dt}) +by=g(t)\ \\ \frac{d^2y}{dt^2}+(\alpha-1)\frac{dy}{dt}+by=g(t)\]

Answer 5

there is a little typo in my first reply second line last term e^(-2t)*****

Answer 6

Fantastic work mukushala ,

Answer 7

tnx my friend

Answer 8

but the typo your referring to is in the last term right?
\[\frac{d^2y}{dx^2}=\cdots=e^{-2t}(\frac{d^2y}{dt^2})-e^{-2t} \frac{dy}{dt}\]

Answer 9

yes

Answer 10

\[\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)={\color\blue{\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}}}\color\red=e^{-t} \frac{d}{dt}\left(e^{-t} \frac{dy}{dt}\right)=e^{-2t}\left(\frac{d^2y}{dt^2}\right)-e^{-2t} \frac{dy}{dt}\]

Answer 11

i dont understand this step,

Answer 12

which part of it?

Answer 13

just this bit
\[{\color\blue{\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}}}\color\red=e^{-t} \frac{d}{dt}\left(e^{-t} \frac{dy}{dt}\right)\]

Answer 14

oh u colored it... sorry...

firstly we proved that \[ \frac{dy}{dx}=e^{-t} \frac{dy}{dt} \]
and we know that
\[ \frac{dt}{dx}=e^{-t} \]

Answer 15

hmm yeah

Answer 16

i think i get it now , thanks again

\[x^2\frac{\text d^2y}{\text d{x^2}}+\alpha x\frac{\text dy}{\text dx}+by=f(x)\]\[\frac{\text dt}{\text dx}=e^{-t}\]
\[\frac{\text dy}{\text dx}=\frac{\text dy}{\text dt} \frac{\text dt}{\text dx}=e^{-t} \frac{\text dy}{\text dt}\]
\[\frac{\text d^2y}{\text dx^2}=\frac{\text d}{\text dx}\left(\frac{\text dy}{\text dx}\right)=\frac{\text d}{\text dx}\frac{\text dt}{\text dt}\left(e^{-t} \frac{\text dy}{\text dt}\right)=\frac{\text dt}{\text dx}\frac{\text d}{\text dt}\left(e^{-t} \frac{\text dy}{\text dt}\right)\]\[\qquad\qquad=e^{-t}\frac{\text d}{\text dt} \left(e^{-t} \frac{\text dy}{\text dt}\right)=e^{-2t}\frac{\text d^2 y}{\text dt^2}-e^{-2t}\frac{\text dy}{\text dt}\]
\[e^{2t} \left(e^{-2t} \frac{\text d^2y}{\text dt^2}-e^{-2t}\frac{\text dy}{\text dt}\right)+\alpha e^{t} \left(e^{-t} \frac{\text dy}{\text dt}\right) +by=g(t) \]
\[ \frac{\text d^2y}{\text dt^2}-\frac{\text dy}{\text dt}+\alpha \frac{\text dy}{\text dt}+by=g(t)\]\[ \frac{\text d^2y}{\text dt^2}+\left(\alpha-1\right)\frac{\text dy}{\text dt}+by=g(t)\]