Question

Solve the De xy''''+4y'''=0. Hint: let v=y'''

Answer 1

\[-3x ^{-3}\]

Answer 2

So far thats what i got after doing the subsitution. But how do i get the equations

Answer 3

Letting \(v=y^{(3)}\) gives \(v'=y^{(4)}\). Subbing into the equation yields
\[xv'+4v=0\]
Rearranging and separating the differentials gives
\[\begin{align*}xv'+4v&=0\\\\
xv'&=-4v\\\\
\frac{1}{v}\frac{dv}{dx}&=-\frac{4}{x}\\\\
\frac{dv}{v}&=-4\frac{dx}{x}\end{align*}\]
Integrate and solve for \(v\), then back substitute.

Answer 4

\[\ln v=-4 \ln x+k1=\ln x^{-4}+k1\]

Answer 5

\[v=k2 x^{-4}=u'\]

Answer 6

\[u=k2 -3x^{-3}+k3\]

Answer 7

Given that
\[v=k_2x^{-4}\]
you would back-sub to get
\[y^{(3)}=k_2x^{-4}\]
and integrate until you can find an expression for \(y\).

Answer 8

So then

\[y^2=-3x^{-3}\]
\[y=\frac{ 3 }{ 2x^2 }\]

Answer 9

Is that the correct way to go about it

Answer 10

Close, but you're missing the constants and a few terms that you'd get from successive integration:
\[\begin{align*}y^{(3)}&=k_2x^{-4}\\
&\Downarrow\\
y''&=\int k_2x^{-4}~dx\\
&=-\frac{k_2}{3}x^{-3}+k_3\\
&\Downarrow\\
y'&=-\frac{k_2}{3}\int x^{-3}~dx+k_3\int dx\\
&=\frac{k_2}{6}x^{-2}+k_3x+k_4\\
&\Downarrow\\
y&=\int\cdots\\
&=-\frac{k_2}{12}x^{-1}+\frac{k_3}{2}x^2+k_4x+k_5
\end{align*}\]

Answer 11

Okay i got this. Thanks. I got one more question. Would you help me

Answer 12

\[2(x-3)^2y''+5(x-3)y'-2y=0\]
Hint: Let
\[t=x-3\]

Answer 13

Using the sub gives
\[2t^2y''+5ty'-2y=0\]
This is an Euler-Cauchy equation, which assumes the solution will be proportional to \(t^r\) for some constant \(r\).
\[y=t^r~~\implies~~y'=rt^{r-1}~~\implies~~y''=r(r-1)t^{r-2}\]
Subbing into the equation gives
\[\begin{align*}2t^2[r(r-1)t^{r-2}]+5t[rt^{r-1}]-2t^r&=0\\\\
2r(r-1)t^r+5rt^r-2t^r&=0\\\\
2r(r-1)+5r-2&=0
\end{align*}\]
Solve for \(r\).

Answer 14

Since the root are \[r=-2, \frac{ 1 }{ 2 }\]
\[y(x)=c1t^{-2}+c2^{\frac{ 1 }{ 2 }}=c1(x-3)^{-2}+c2(x-3)^{\frac{ 1 }{ 2 }}\]