# Learning Differential equations, week 2 level material

Tags:
1. Oct 1, 2015

### SYoungblood

1. The problem statement, all variables and given/known data

Find the general solution to the equation.

2. Relevant equations
(dy/dx) - y - e^3x=0

3. The attempt at a solution

I rewrote this as dy/dx - y = e^3x

This is a linear first order ODE, in the form dy/dx + P(x)y = f(x)

P(x) = 1; f(x) = e^3x

The integrating factor = e^(integral)P(x) dx = e^(integral) 1* dx = e^x

Set the LHS equal to the integrating factor, e^x * y, and multiply the RHS by the integrating factor e^x)

d/dx (ye^x) = e^x e^3x

(integral) d/dx (ye^x) =(integral) e^(3x^2) dx

On the LHS, the integral and d/dx cancel out, but I am getting a ridiculously complicated integral that I just know isn't right on the RHS, and I am not sure where I went astray.

SY

2. Oct 1, 2015

### HallsofIvy

Staff Emeritus
$e^xe^{3x}= e^{4x}$, NOT $e^{3x^2}$.

3. Oct 1, 2015

### Staff: Mentor

Please use parentheses where needed. e^3x would normally be interpreted as $e^3 \cdot x$. In your first equation above, you have parentheses around dy/dx, where they aren't needed, but don't have them around the exponent 3x, where they are needed.

BTW, I changed the thread title to say "... week 2 level material". I think that's what you intended.

4. Oct 1, 2015

### SYoungblood

will do, thank you

5. Oct 1, 2015

### SYoungblood

Thank you, that is not the first silly mistake I will make, and will not be the last...