The differential equation $\displaystyle y' + 2xy = 4e$ on $\displaystyle -x^2$ is given.
Solve the modified homogeneous equation.
What is the general solution to the equation?
Find that particulate solution which goes through T(0,2).
The differential equation $\displaystyle y' + 2xy = 4e$ on $\displaystyle -x^2$ is given.
Solve the modified homogeneous equation.
What is the general solution to the equation?
Find that particulate solution which goes through T(0,2).
First, I assume you ment $\displaystyle y'+2xy=4e^{-x^2} $.
Anyway, the equation is a first order linear differential equation. In general it can be written down as $\displaystyle y' + p(x)y=q(x)$. Using the variation of the constant method general solution formula can be found, and it goes something like this $\displaystyle y=\left(\int q(x)\cdot e^{-P(x)} dx+C\right)\cdot e^{P(x)}$ where $\displaystyle P$ is the primitive function of the function $\displaystyle p$, $\displaystyle \qquad \int p(x) dx= P(x)+const.$.
Using the formula should suffice to solve the problem(s).