I'm trying to solve an initial value problem. I know how to do the problem once integrated but solving the differential equation is where I'm finding trouble.
$$dy/dx = 3 +\sqrt {2y+17x-3}$$
I'd thought that maybe squaring both sides will get rid of the square root but that doesn't work so I was hoping someone would point me in the right direction of how to go about seperating $y$ and $x.$
On
$$\frac{dy}{dx} = 3 +\sqrt {2y+17x-3}$$ Let $\quad 2y+17x-3=u^2\quad$ with $\quad u\geq 0\quad\to\quad y'=-\frac{17}{2}+u'u=3+u$
$u'u=u+\frac{23}{2}$ $$x=\int \frac{udu}{u+\frac{23}{2}}=u-\frac{23}{2}\ln(u+\frac{23}{2})+c$$ This gives $x(u)$. The inverse function $u(x)$ involves a special function : $$u=\frac{23}{2}\left(W(X)-1\right)\quad \text{where}\quad X=\frac{2}{23}\exp(1+2\frac{x-c}{23})$$ $W(X)$ is the Lambert's W function : http://mathworld.wolfram.com/LambertW-Function.html $$\begin{cases}y(x)=-\frac{17x+3}{2}+\frac12\left(\frac{23}{2}\left(W(X)-1\right) \right)^2 \\ X=\frac{2}{23}\exp(1+2\frac{x-c}{23})\end{cases}$$
The differential equation $$dy/dx = 3 +\sqrt {2y+17x-3}$$
is not separable but a change of variable turns it into a separable equation.
Let $$ u=2y+17x-3$$
Then $$du/dx = 2dy/dx +17$$
Substitute in the original equation to get $$du/dx =2 \sqrt u +23$$
Which is separable.
Solve for $u$ and back substitute your solution to find $ y$ as a function of $x$