How do I solve a first order differential equation of the following form If I had a differential equation $x'(t) + \frac{x(t)}{t} = e^{t^2}$
And I have a question saying solve the following first ODE for $x(t)$ how would I go about doing this?
 A: This is a linear differential equation so first let's find the solution to the homogenous equation. We see that $x(t)=A\exp(-\ln(t))=\frac{A}{t}$ where $A$ is a constant. Using a well-known method, we suppose that $A=A(t)$, so we have $x(t)=\frac{A(t)}{t}$. Using the equation, we get$$
A'(t)=t \exp(t^2)
$$
so $A(t) = \frac{\exp(t^2)}{2}+C$ where $C$ is a constant. Thus
$$
x(t)=\frac{\exp(t^2)}{2t}+\frac{C}{t}.
$$
A: With this kind of problems, the standard approach is to start with finding a solution to the homogenous differential equation, and then add what's called a 'particular solution' for the inhomogenous part. If you're not familiar with this solving method, you should learn more about it very soon, it's quite common :)
For the homogenous equation:
$x'(t) = -\frac{x(t)}{t}$
I used the so called 'separation of the variables' method:
$\frac{dx}{x(t)} = -\frac{dt}{t}$
And came to the homogenous solution:
$x(t) = \frac{c}{t}$
for some constant $c \in \mathbb{R}$.
Then, to find a particular solution (you only need one of them), I used the method of the variation of constants:
$x(t) = \frac{1}{t}\left(c + \int_{0}^{t}x\cdot e^{x^2}dx\right)$
And finally found the following general solution to your ODE problem:
$x(t) = \frac{1}{2}\cdot\frac{e^{t^2}}{t} + \frac{c}{t}$
So, I hope this will help you to understand the steps to follow! And I wrote my results so that you can check on your way :)
A: $$x'(t) + \frac{x(t)}{t} = e^{t^2}\tag1$$
It is a first order linear ODE. Integrating factor $~=e^{\int(1/t)dt}=e^{\ln t}=t~.$
Multiplying both side of $(1)$ with the integrating factor, we have
$$tx'(t) + x(t)= te^{t^2}\tag2$$Integrating,
$$\implies d\left(tx\right)=\int te^{t^2}~dt$$
$$\implies tx=\dfrac 12\int e^{t^2}~d(t^2)$$
$$\implies tx=\dfrac 12 e^{t^2}+c$$
$$\implies x(t)=\dfrac {e^{t^2}}{2t}+\dfrac{c}{t}$$where $c$ is integrating constant.
