Question

Solve the initial-value problem y'-xy=xy^(3/2), y(1)=4.

Answer 1

@SithsAndGiggles

Answer 2

\[\begin{align*}
y'-xy&=xy^{3/2}\\
y'&=xy+xy^{3/2}\\
y'&=xy(1+y^{1/2})\\
\frac{dy}{y(1+\sqrt y)}&=x~dx
\end{align*}\]
Is this what you have so far?

Answer 3

I don't have that so far because I did the Bernoulli method. So can I do it the way you did instead of Bernoulli?

Answer 4

Yes, the equation is separable. I'm not sure the Bernoulli substitution works for rational values of \(n\). Let's check...
\[\begin{align*}
y'-xy&=xy^{3/2}\\
y^{-3/2}y'-xy^{-1/2}&=x
\end{align*}\]
So we'd use the substitution, \(v=y^{-1/2}\), which gives \(v'=-\dfrac{1}{2}y^{-3/2}y'\), so
\[\begin{align*}
2v'-xv&=x\\
v'-\frac{x}{2}v&=\frac{x}{2}
\end{align*}\]
I guess it's still valid! Find your integrating factor and you'll be set.

Answer 5

I got v'-x/2*v=-x/2 instead of x/2 like yours. And the integrating factor I got is e^(x^2/4) but how do I integrate (x/2)(e^(x^2/4))?

Answer 6

Oh yeah that's a typo on my part, sorry. Should be
\[\begin{align*}
\color{red}{-2}v'-xv&=x\\
v'+\frac{x}{2}v&=-\frac{x}{2}
\end{align*}\]

Answer 7

So do I do integration by parts to integrate (-x/2)(e^(x^2/4))?

Answer 8

\[\mu(x)=\exp\left(\int\frac{x}{2}~dx\right)=\exp\left(\frac{1}{4}x^2\right)=e^{(x^2)/4}\]
\[\begin{align*}
e^{(x^2)/4}v'+e^{(x^2)/4}\frac{x}{2}v&=-e^{(x^2)/4}\frac{x}{2}\\
\frac{d}{dx}\left[e^{(x^2)/4}v\right]&=-e^{(x^2)/4}\frac{x}{2}\\
e^{(x^2)/4}v&=-\frac{1}{2}\int xe^{(x^2)/4}~dx
\end{align*}\]
IBP isn't necessary here. Just substitute \(t=\dfrac{x^2}{4}\), so \(dt=\dfrac{x}{2}~dx\), or \(2~dt=x~dx\).
\[\begin{align*}e^{(x^2)/4}v&=-\int e^t~dt\\
e^{(x^2)/4}v&=-e^t+C\\
e^{(x^2)/4}v&=-e^{(x^2)/4}+C\\
v&=-1+Ce^{-(x^2)/4}
\end{align*}\]

Answer 9

Let me work it out.

Answer 10

So y=(-1+(e^(x^2/4))/C)^2, right?

Answer 11

We used \(v=y^{-1/2}\), which means \(y=\dfrac{1}{v^2}\). You should then have
\[y=\frac{1}{\left(Ce^{-(x^2)/4}-1\right)^2}\]

Answer 12

Got it! Thanks!

Answer 13

You're welcome!

Answer 14

:)