Integrating Factor problem

[rippletank]

Solving the following problem for a general solution, I cannot understand how you get from the first step to the second.

d/dt(e^t^2) + 2ty(e^t^2) = 4e^-t^2(e^t^2)

d/dt(ye^t^2) = 4

I understand the right hand side, but I don't understand how the left hand side of the first line can be simplified to the left hand side of the second. That is what the book gives, but I must be missing something. Thanks for any help.

The original problem is
dy/dt = -2ty + 4e^-t^2 , y(0)=3

I know the integrating factor is e^t^2

 

[Deleted member 4993]

Solving the following problem for a general solution, I cannot understand how you get from the first step to the second.

dy/dt * (e^t^2) + 2ty(e^t^2) = 4e^-t^2(e^t^2)

d/dt(ye^t^2) = 4

I understand the right hand side, but I don't understand how the left hand side of the first line can be simplified to the left hand side of the second. That is what the book gives, but I must be missing something. Thanks for any help.

The original problem is
dy/dt = -2ty + 4e^-t^2 , y(0)=3

I know the integrating factor is e^t^2

d/dt [y * e^(t^2)] = d/dt (y) * e^(t^2) + y * d/dt [e^(t^2)] ... now continue....

 

[HallsofIvy]

An integrating factor for an expression such as $\displaystyle f(t) dy/dt+ g(t)y$ is a function, $\displaystyle \mu(t)$, such that $\displaystyle d(\mu(t) y)= \mu(t)f(t)dy/dx+ \mu(t)g(t)y$. That is, multiplying the expresson by it converts it to a single derivative. By the product rule, $\displaystyle d(\mu(t)y)/dt= \mu(t) dy/dt+ \mu'(t)y$ and, for this problem, that must be equal to $\displaystyle \mu(t)dy/dx+ 2\mu(t)ty$

$\displaystyle \mu(t) dy/dt+ \mu'(t)y= \mu(t) dy/dt+ 2t\mu(t)y$

The "$\displaystyle \mu(t) dy/dt$" terms cancel so we must have $\displaystyle \mu'(t)= 2t\mu(t)$ which is separable: $\displaystyle \frac{d\mu}{\mu}= 2t dt$. Integrating $\displaystyle ln(\mu(t))= t^2+ C$ so that $\displaystyle \mu(t)= C' e^{t^2}$. Since we only need one such function, take C'= 1 which is equivalent to taking C= 0.

That is, by the product rule, $\displaystyle (e^{t^2}y)'= e^{t^2}y'+ 2te^{t^2}y$. Multiplying the original equation, \(\displaystyle dy/dt+ 2\ty= 4e^{-t^2}\) by $\displaystyle e^{t^2}$ gives

$\displaystyle e^{t^2}dy/dt+ 2te^{t^2}y= d(e^{t^2}y)/dt= 1$