Question

Solve dy/dx - XY = xy^3​

Answer 1

y′+y=xy3

The above equation is known as Bernoulli equation. It’s general form is:

y′+p(x)y=q(x)ym

In our case “p(x)” is 1 and “q(x)” is x . The value of m is 3.

In this case, one way for solving such equation is through substitution:

v=y1−m → v=y−2

After a few “algebraic manipulations”, we get an equation that looks like this:

v′+(1−m)p(x)v=(1−m)q(x) →

v′−2v=−2x

Which is a linear non-homogeneous equation!

First, let’s find ourselves an integrating factor:

u(x)=e∫−2dx=e−2x

Multiply both sides:

u(x)v′−2vu(x)=−2xu(x) →

e−2xv′−2ve−2x=−2xe−2x

Next, you may notice that on the left hand side there is an expression that reminds of derivative product rule:

(uv)′=u′v+uv′

Therefore, we can write it like this:

(ve−2x)′=−2xe−2x

Now, let’s integrate both sides:

∫(ve−2x)′dx=∫−2xe−2xdx

The integral on the left hand side cancels out with the derivative, and we are left with the right hand integral.

∫−2xe−2xdx

To solve it, we will use integration by parts:

∫−2xe−2xdx=[v′=e−2xv=−12e−2xu=−2xu′=−2]=xe−2x−∫e−2xdx

xe−2x+12e−2x+c

The total solution is:

ve−2x=xe−2x+12e−2x+c

v=x+12+e2xc

v=2x+1+2e2xc2

The last step is to go back y:

1y2=2x+1+2e2xc2

y2=22x+1+2e2xc→y=±22x+1+2e2xc−−−−−−−−√

Therefore, the general solution is:

y=±22x+1+2e2xc−−−−−−−−√

One more thing…. Since some of “algebraic manipulations” mentioned above involved dividing by “y”,

We now must check whether y≡0 is a solution.

By plugging it into our equation, we get:

0+0=x0→0=0

That means that y ≡0 may be a private or special solution.

However, by plugging it into our general solution we get:

0=±22x+1+2e2xc−−−−−−−−√→0=2→0≠2

Which means that y ≡0 is not a private, but a singular/special solution.