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 $x.$

  • $\begingroup$ Bring 3 to the other side and square then? $\endgroup$ – user370967 Mar 11 '18 at 15:17

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 $x$

  • $\begingroup$ I'm stuck at the last step with solving for u and back substituting. Does that mean seperate and integrate then sub 2y+17x−3 back in? $\endgroup$ – OrangeSegments Mar 11 '18 at 15:48
  • $\begingroup$ Yes, that is what I meant. Solve for u first. and then sub 2y+17x−3 back in $\endgroup$ – Mohammad Riazi-Kermani Mar 11 '18 at 15:52

$$\frac{dy}{dx} = 3 +\sqrt {2y+17x-3}$$ 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 $$\begin{cases}y(x)=-\frac{17x+3}{2}+\frac12\left(\frac{23}{2}\left(W(X)-1\right) \right)^2 \\ X=\frac{2}{23}\exp(1+2\frac{x-c}{23})\end{cases}$$


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