# DE#8

#### idontknow

(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 $$\displaystyle y(x)=\int_{0}^{1} (1-t^2 )^\lambda \cos(xt)dt$$.

#### idontknow

(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) ?

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#### skipjack

Forum Staff
(3) If $c = 0$, just integrate, else use $\cos(cx)$ as an integrating factor.

#### skipjack

Forum Staff
(2) $$\displaystyle n\in\mathbb{N}$$ and should I set $$\displaystyle x=e^t$$ ?
Do you mean that $n\in\mathbb{N}$ should have been stated originally?
Let $y = ze^x$. You should get xy'' + (1 - x)y' + ny = 0, which is the Laguerre Differential Equation.

#### idontknow

got it .
(4) asks just to substitute the given solution .

#### topsquark

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

-Dan

#### skipjack

Forum Staff
(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.

#### idontknow

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.

#### skipjack

Forum Staff
After finding $z$, you still need to integrate it correctly to solve $yy' = z$.