Please help me How to solve the equation y'=-x^2-sin(y)

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Question the problem:

y'=-x^2-sin(y)

please solve the question
thank you
 
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We are willing to help you solve the problem - we are not here to solve it for you.
 
SammyS said:
What have you tried?

Where are you stuck?
i live in Thailand

We are willing to help you solve the problem - we are not here to solve it for you.

i solve in ODE
step 1. set question dy/dx+P(x)y=g(x)
step 2. P(x)=exp(integrate(F(x))dx)
step 3. x p(x) in 1
but
----->y'=-x^2-sin(y)
----->y'+sin(y)=-x^2 ----1
------>exp(integrate(1)dx
------>exp(x) ----2
x exp(x) in 1

exp(x)y'+exp(x)sin(y)=exp(x)*-x^2 I made ​​up this process safe.
It was not possible to do.ฃ
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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