# Solving a differential-initial value equation: $y'(x) = \frac{y}{x+y^3}$, $y(2)=2$

The equation and given values are $$\begin{split} \frac{dy}{dx} &= \frac{y}{x+y^3}\\ y(2) &= 2 \end{split}$$

At first I thought it was a separation of variables problem, but I then was told later on by a tutor that it could be solved with substitution. I then subbed-in with the following variables: $$u+x\frac{du}{dx}=\frac{ux}{x+(ux)^3}$$
Then subtracted the $$u$$ from both sides and multiplied by the denominator: $$x\frac{du}{dx}=\frac{ux-u(x+(ux)^3)}{x+(ux)^3}$$ $$x\frac{du}{dx}=\frac{ux-ux-u^4x^3}{x+u^3x^3}$$ $$x\frac{du}{dx}=\frac{-u^4x^3}{x+u^3x^3}$$ Now If I factor out an x from the quotient $$x\frac{du}{dx}=-\frac{x(u^4x^2)}{x(1+u^3x^3)}$$ $$x\frac{du}{dx}=-\frac{u^4x^2}{1+u^3x^3}$$ Now can I further simplify the quotient till I can integrate or should I go back and do something different to make it easier.

It's not a homogeneous differential equation to use that substitution $$(y=tx)$$. You can consider $$x'=\dfrac {dx}{dy}$$ instead of $$y'$$, then the DE becomes linear. Or you can try this : $$\frac{dy}{dx} = \frac{y}{x+y^3}\\$$ $$({x+y^3}){dy} = {y}{dx}$$ $${y}{dx}-x{dy}=y^3{dy}$$ $$\frac {{y}{dx}-x{dy}}{y^2}=y{dy}$$ $$d \left ( \frac x y\right )=y{dy}$$ After integration it gives: $$\frac x y =\dfrac 12 y^2 +C$$ $$x$$ as a function of $$y$$: $$x(y) =\dfrac 12 y^3 +Cy$$ Apply initial condition: $$y(2)=2 \implies C=-1$$ Finally: $$\boxed { x(y) =\dfrac { y^3}2 -y}$$

• What did you do on the 5th line? – Eric Brown Feb 18 at 20:12
• @EricBrown Thats a well known result $$d \left (\frac xy \right ) = \dfrac {ydx-xdy}{y^2}$$ – LostInSpace Feb 18 at 20:14
• @EricBrown Thats easy Eric remember the rule $$\int d(something)=something +C$$ Got it ? – LostInSpace Feb 18 at 20:33
• Yeah I figured it out – Eric Brown Feb 18 at 20:35
• @EricBrown That's great ... – LostInSpace Feb 18 at 20:36

HINT Invert both sides of the equation to make it a linear first order equation and solve.

ANSWER After inverting we get, $$\frac{dx}{dy}=\frac{x}{y}+y^2,$$ which equates to :$$\frac{dx}{dy}-\frac{x}{y}=y^2.$$ Multiply both of the equation with $$e^{\int \frac{-1}{y}dy}=\frac{1}{y}$$.

Integrating the total derivative we then get $$\frac{x}{y}=\int ydy,$$ which equates to $$x(y)=\frac{y^3}{2}+cy,$$ for some integration constant $$c$$.

Solve with the initial conditions to get the answer from there, which is that $$c=-1$$ and so $$x(y)=\frac{y^3}{2}-y,$$

• How did you multiply out the integration factor? – Eric Brown Feb 18 at 19:57

Hint:

Moving the denominator, we have

$$xy'+y^3y'=y.$$

Then we notice the terms $$y-xy'$$ which can be the numerator of the derivative of $$\dfrac xy$$. So we rewrite

$$\frac{y-xy'}{y^2}=yy'$$ or

$$\left(\frac xy\right)'=\frac12(y^2)'$$

The differential equation, $$\frac{dy}{dx}=\frac{y}{x+y^3}$$ is not homogenous so the substitution $$y=ux$$ is not helpful.

If you change your eqaution into $$\frac{dx}{dy}=\frac{x+y^3}{y}$$

then it is linear and you may solve it using integrating factor method.

$$\frac{dx}{dy}=\frac{x+y^3}{y}=(1/y)x+y^2$$

The integration factor is $$1/y$$ and after multipluing by $$1/y$$ and integrating you get the solution $$x=\frac {1}{2} y^3 +cy$$

The initial condition of $$y(2)=2$$ gives you $$x=\frac {1}{2} y^3-y$$