Question

Find a curve y=y(x) through (1, -1) such that the tangent to the curve at any point (xo, y(xo)) intersects the y axis at yI=(xo)^3.

Answer 1

Let me show you my work first.

Answer 2

Wasn't this already asked? http://openstudy.com/users/idealist10#/updates/54e3a0cbe4b0f2455c588421

Now you're given an explicit initial condition.

Answer 3

\[y-y(x _{0})=m(x-x _{0})\]

Answer 4

Oh it's y-axis this time...

Answer 5

(0, (xo)^3)
(xo)^3-y(xo)=m(0-xo)
(xo)^3-y(xo)=-m(xo)
-m=((xo)^3-y(xo))/(xo)
m=(y(xo)-(xo)^3)/(xo) at (xo, y(xo))
m=dy/dx=(y-x^3)/x
dy/(y-x^3)=dx/x
And I have trouble integrating dy/(y-x^3).

Answer 6

You might have better luck treating this as a linear equation. It doesn't appear to be separable.
\[\begin{align*}
{x_0}^3-y_0&=y'(x_0)(0-x_0)\\\\
-{x_0}^2&=y'(x_0)-\frac{1}{x_0}y_0\end{align*}\]

Answer 7

How did you get -(xo)^2=y'(xo)-(1/xo)(yo)?

Answer 8

\[\begin{align*}
{x_0}^3-y_0&=y'(x_0)(0-x_0)\\\\
y_0-{x_0}^3&=y'(x_0)x_0\\\\
-{x_0}^3&=y'(x_0)x_0-y_0\\\\
-{x_0}^2&=y'(x_0)-\frac{1}{x_0}y_0\end{align*}\]

Answer 9

So from there, what do you do?

Answer 10

The ODE is linear, so you can go ahead with determining the integrating factor.

Answer 11

Got it! Thanks!