Solve (3ycosx+4xe^x+2x^2*e^x)dx+(3sinx+3)dy=0.

Question

idealist10 · Answer

@amistre64 @hartnn @dumbcow

idealist10 · Answer

Please help me. This is an exact differential equation.

amistre64 · Answer

if its exact  then integrate one side, then derive it to the other side, is what i recolect

idealist10 · Answer

Can you show your work, amistre?

dumbcow · Answer

ok i dont have a lot of time to work through this, i will get you started http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx $$\int\limits N dy = \int\limits (3 \sin x +3) dy = C$$ $$3 y \sin x +3y +h(y) = C$$

amistre64 · Answer

int: (3+3sinx) dy = 3y-3ysin(x)+f(x)

Dx: 3y-3ysin(x)+f(x) = -3ycos(x)+f'(x)

now compare

-3ycos(x)+f'(x) = 3ycosx+4xe^x+2x^2*e^x

amistre64 · Answer

the sites starting to lag ... bummer

amistre64 · Answer

i see i fubarred a neative

amistre64 · Answer

3ycos(x)+f'(x) = 3ycosx+4xe^x+2x^2*e^x

f'(x) = 4xe^x+2x^2*e^x

dumbcow · Answer

oops i meant h(x)
then differentiate with respect to x
$$\rightarrow 3y \cos x +h'(x) = \frac{d}{dx}(3 \sin x +3) = 3 \cos x$$

$$h'(x) = 3 \cos x (1-y)$$

amistre64 · Answer

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