0
$\begingroup$

I this trying to solve y'' = yy' with y(0) = 1, y'(0) = 1.

I know y' = 1/x', y'' = -x''/(x')^3 . Then I tried substitute u = x'. So du = x''. y'' = -du/u^3. Then \int y'' = \frac{1}{2u^2}. I am a bit confused at this point. Could someone point out how to proceed?

  • 0
    @GerryMyerson: yes2011-10-13

1 Answers 1

1

Even though this can be done just by integrating both sides, the following technique might be of interest.. sometimes for such equations it helps to write y'(x) = z(y(x)). Taking derivatives you get y''(x) = z'(y(x)) y'(x) = z'(y) z . So the equation y'' = y y' becomes, when viewed as functions of $y$: z'(y) z = yz Which is the same as z'(y) = y So $z(y) = {1 \over 2}y^2 + C$ So you are reduced to solving ${dy \over dx} = {1 \over 2}y^2 + C$ You can separate variables to solve this.