ODE with separable variables and three conditions

Question

I have to solve $y^{'}(x)=cos(x)y^{3}(x)$.

Seemingly straightforward, I write $$\displaystyle \int \frac{dy}{y^3}=\int \cos x dx\rightarrow \frac{y^{-3+1}}{-3+1}=\sin x+c$$

$$ -\frac{1}{2}y^{-2}=\sin x+c\rightarrow y^2=\frac{1}{-2(\sin x+c)} \Rightarrow y=\pm \left(-\frac{1}{\sqrt{2}\sqrt{c+\sin x}} \right)$$

Now the text asks to find the solution "with the initial condition $y(x_0)=y_0$ in the three different cases:

• $y_0=0$
• $y_0>0$
• $y_0<0$

This is the first time i have come across a problem of this nature, i don't know how to solve it. Am i still create the system?

Thanks for any help!

Answer 1

The solution is real since $y_0$ is real.

• The case $y_0=0$ was solved by @Kavi Rama.

The general non-zero solution verifies $y^2=\frac{-1}{2(\sin x +c)},$ and is real only if $(\sin x + c)$ is negative.

• $y_0>0$ leads to $y_0=\sqrt\frac{-1}{2(\sin 0 +c)}=\sqrt{\frac{-1}{2c}},$ from where $c=-\frac{1}{2{y_0}^2}$ and therefore $y=\sqrt\frac{-1}{2(\sin x -1/(2{y_0}^2))}.$
• Similarly, $y_0<0$ leads to $y_0=-\sqrt{\frac{-1}{2c}}.$ Again, $c=-\frac{1}{2{y_0}^2}$ and $y=-\sqrt\frac{-1}{2(\sin x -1/(2{y_0}^2))}.$

Answer 2

The only solution in the first case is $y=0$. (Because when you assumed that $y \neq 0$ and divided by $y^{3}$ you got a solution which does not satisfy $y_0=0$. Let $f(x)$ be the integral of $\sec x$. For $y_0 >0$ the solution is $\frac 1 {\sqrt 2 \sqrt {c+ f(x)}}$ and For $y_0 <0$ the solution is $-\frac 1 {\sqrt 2 \sqrt {c+f(x)}}.$