Integration by separation of variables

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Homework Statement



Using the technique involving variable separation, solve the following differential equation and use the initial condition to find the particular solution

\frac{dy}{dt} = \frac{1}{y^{2}} y(0) = 1

Homework Equations






The Attempt at a Solution



To be honest, could someone show me how to solve this one as I cannot seem to understand how to get to the book answer of y=(3t+1)^{1/3} .

I understand that the y^2 should move to the left so that it becomes dy/y^2 but I am not sure how to progress with this?
 
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Physicist3 said:

Homework Statement



Using the technique involving variable separation, solve the following differential equation and use the initial condition to find the particular solution

\frac{dy}{dt} = \frac{1}{y^{2}} y(0) = 1

Homework Equations






The Attempt at a Solution



To be honest, could someone show me how to solve this one as I cannot seem to understand how to get to the book answer of y=(3t+1)^{1/3} .

I understand that the y^2 should move to the left so that it becomes dy/y^2 but I am not sure how to progress with this?
The y2 should be moved to the left side, but it doesn't become dy/y2.

What operation do you need to apply to both sides to get rid of the y2 in the denominator on the right?
 
You should separate the variables, first move to the left the squared term, then move dt, now you must integrate don't forget the constant, now solve the equation for y. Y(0) it's a initial condition so when y = 0, t = 1, you should get a function with these conditions and this function ought to be the solution of the ODE.
 
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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