How to solve the differential equation with substitution?

Question

y'=(x^2-y^2)xy, with y(1)=2

Answer 1

y'=(x^2-y^2)xy, with y(1)=2

 

First equation.

y'=x³y-xy³ (This is, so called, Bernoulli equation)

y'-x³y = -xy³ (multiply both sides by y^-3)

y^-3y'-x³y^-2=-x

Now we make substitution u=y^-2, where u=u(x) is a new unknown function. Derivative of u: u'=-2y^-3y',    -u'/2=y^-3y'.

Therefore

-u'/2-x³u=-x

or 

u'+2x³u=2x  (this is first order linear equation)  (**)

We solve this equation and then go back to y

First this : u'+2x³u=0. If you solve this you get u=C e^(-x⁴/4), C=const

Assume that C=C(x) (that is, for time-being not constant but function of x) and substitute into equation (**).

If you do that, you find 

C(x)= C₀-e^(-x⁴/4)  and 

u(x)=C₀e^(-x⁴/4)-e^(-x⁴/2).

remind that u=y^-2

So

y^-2=C₀-e^(-x₄/4)-e^(-x⁴/2).

y=1/(C₀e^(-x⁴/4)-e^(-x⁴/2))^(1/2)

 

You have y(1)=2. So

1/4=C₀- e^(-1/4)-e^(-1/2)

C₀=1/4+ e^(-1/4)-e^(-1/2)

 

Plug in C₀ and you have that solution with initial condition y(1)=2.

Note:I could made mistakes in details. If you go through every step yourself, you might find inconsistencies (if any). But, say, the approach is correct.

Of course you can solve it numerically as suggested Andy C. Numerical solution is good. This is in addition to that.

 

 

Answer 2

You can use a numerical method like Runge Kutta