Solving a Simple Differential Equation: y' = (x + xy^2)

S[e^x]=f(u)^n
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Homework Statement


i'm trying to solve the simple differential equation y'=(x+xy^2)


2. The attempt at a solution

i get dy/(1+y^2)=xdx and then integrate over the whole thing getting the solution

arctan(y)=0.5x^2+C

my problem is how do i simply this down to an equation in the form y=...

i must be missing something
 
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S[e^x]=f(u)^n;1635559 said:

Homework Statement


i'm trying to solve the simple differential equation y'=(x+xy^2)


2. The attempt at a solution

i get dy/(1+y^2)=xdx and then integrate over the whole thing getting the solution

arctan(y)=0.5x^2+C

my problem is how do i simply this down to an equation in the form y=...

i must be missing something
well remember that if tan(x)=y then arctan(y)=x does this help u? Tan and arctan are inverse functions so, tan(arctan x)=x
 
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but that x only valid under certain bounds right? and I'm having trouble figuring out what those are
 
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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