DE#8

Dec 2015
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Earth
(1) \(\displaystyle yy''+y'^2 =\frac{yy'}{\sqrt{1+x^2 }}\).

(2) \(\displaystyle x\frac{d^2 z}{dx^2 }+(x+1)\frac{dz}{dx}+(n+1)z=0\).

(3) \(\displaystyle \frac{d^2 y}{dx^2 }+c^2 y =\phi(x) \: , c\in \mathbb{R}\:\) .

(4) If \(\displaystyle xy''+2(\lambda +1 )y' +xy =0 \: , \: \lambda>-1\) , prove \(\displaystyle y(x)=\int_{0}^{1} (1-t^2 )^\lambda \cos(xt)dt\).
 
Dec 2015
1,085
169
Earth
(1) \(\displaystyle (yy')'-\frac{yy'}{\sqrt{1+x^2 }} =0 \) ; \(\displaystyle \; \; \ln yy' -\int \frac{dx}{\sqrt{1+x^2}}=c_1\).

\(\displaystyle y=c_1[ x^2 + x\sqrt{1+x^2 }+\ln (x+\sqrt{1+x^2 })] +c_2\).

(2) \(\displaystyle n\in\mathbb{N}\) and should I set \(\displaystyle x=e^t \) ?

How to solve (3) ?
 
Last edited:
Dec 2015
1,085
169
Earth
got it .
(4) asks just to substitute the given solution .
thread done .
 

topsquark

Math Team
May 2013
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The Astral plane
(3) If $c = 0$, just integrate, else use $\cos(cx)$ as an integrating factor.
Now that's just cool. :cool: I never would have thought of the integrating factor.

-Dan
 

skipjack

Forum Staff
Dec 2006
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(1) \(\displaystyle (yy')'-\frac{yy'}{\sqrt{1+x^2 }} =0 \) ; \(\displaystyle \; \; \ln yy' -\int \frac{dx}{\sqrt{1+x^2}}=c_1\).

\(\displaystyle y=c_1[ x^2 + x\sqrt{1+x^2 }+\ln (x+\sqrt{1+x^2 })] +c_2\).
That's not quite right.
 
Dec 2015
1,085
169
Earth
That's not quite right.
Maybe it must be \(\displaystyle e^{c_1 } \) instead of \(\displaystyle c_1\).
\(\displaystyle yy'=z\) ; \(\displaystyle \; \; z'-\frac{z}{\sqrt{1+x^2 }}=0\) is solved with integrating factor.