Question

Find the general solution of: y'+4x=cos2x
Can you please show me the steps you took to get the answer of this problem.

Answer 1

y'=cos(2x)-4x
just integrate both sides to find y

Answer 2

\[y=\frac{1}{2}\sin(2x)-4\frac{x^2}{2}+C\]

Answer 3

dy/dx = cos2x - 4x
dy = cos(2x) - 4x dx

integrate both sides :

y = sin(2x) / 2 - 2x^2 + C

Answer 4

Oh! you guys are awesome, so every problem like this will involve integrating, right? Okay, I have one more:

Find the general solution of: 4xy dx=(x^2+4)dy

Answer 5

separate your variables
then integrate

Answer 6

\[\frac{4x}{x^2+4} dx=\frac{1}{y} dy\]

Answer 7

I just want to double check on your original problem. It is
y'+4x=cos2x
NOT
y'+4y=cos2x
Yes?

Answer 8

Yes

Answer 9

great