Solving the differential equation $2y'\sin x + y\cos x = y^3(x\cos x - \sin x) $

Question

Does anyone know how to solve the following differential equation:

$$2y'\sin x + y\cos x = y^3(x\cos x - \sin x) $$

I tried dividing both sides by sine, then by cosine, which in either case brought me nowhere. I tried isolating $y'$ but only ended up getting a complex expression which is seemingly impossible to integrate. Are there any other approaches here?

Answer 1

Hint

Let $y=\frac 1 {\sqrt z}$, replace and simplify to get $$z'\sin (x) -z \cos (x)=\sin (x)-x \cos (x)$$ which looks much more pleasant to work.

I am sure that you can take it from here.

Answer 2

Hint

Substitute $u=\frac 1 {y^2}$

Then $$u'=-2 \frac {y'} {y^3}$$ $$2y'\sin x + y\cos x = y^3(x\cos x - \sin x)$$ $$-u' + u\cot (x) = (x\cot(x) - 1)$$ $$ {u'} -u\cot (x) = (1-x\cot(x) )$$ $$.............$$

Answer 3

HINT.

This is a Bernoulli differential equation, that is an ODE of the form

$$y'(x) + p(x) y(x) = q(x) y^n(x)$$

Where $p(x)$ and $q(x)$ are functions.

You can easily transform yours into a Bernoulli type, and the next step will be to use this substitution:

$$z = y^{1-n}$$

In your case, $n = 3$.

After this substitution, your ODE will turn into the form

$$z'(x) + (1-n)p(x) z(x) = (1-n)q(x)$$

Which is then a simple first order linear differential equation, which has the form

$$z'(x) + Q(x)z(x) = f(x)$$

it's straightforward to connect the coefficient to your terms.

Remember then, that the solution of the above ODE is given by

$$\large z(x) = \frac{1}{e^{\int Q(x)\ dx}}\left(\int e^{\int Q(x)\ dx}\ f(x)\ dx + C\right)$$