Differential Equation

Question

itz_sid · Answer

$$y' = 3x-y$$
$$I(x) = e^{\int\limits dx} = e^x$$

itz_sid · Answer

$$\int\limits [e^ty]' = \int\limits e^x(3x)$$

perl · Answer

$$ y' +y = 3x$$

itz_sid · Answer

$$e^ty= \int\limits e^x(3x) dx$$

itz_sid · Answer

And for that integral i tried solving with Integration by parts, but get stuck...

itz_sid · Answer

$$u = x \rightarrow du = \frac{ 1 }{ x }$$
$$dv = e^xdx \rightarrow v=e^x$$

$$ye^x = 3[xe^x- \int\limits \frac{ e^x }{ x }]$$

itz_sid · Answer

Do I have to do IBP again?

nnesha · Answer

derivative of x is ??

itz_sid · Answer

Oh shoot. Thanks haha xD

perl · Answer

$$ y' +1y = 3x
\  ~ \
e^{\int 1 dx} ( y' +1y) = e^{\int 1 dx} 3x
\  ~ \
e^{x} ( y' +1y) = e^{x} 3x
\  ~ \
e^{x}y' +e^{x}y = 3x e^{x} 
\  ~ \
(e^{x}y)'  = 3x e^{x} 
\  ~ \
\int (e^{x}y)'~dx  = \int 3x e^{x} ~dx
\  ~ \
e^{x}y  = \int 3x e^{x} ~dx
\ ~ \ 
e^{x}y  = 3xe^x -3e^x +c
\ ~ \ 
y  = \frac{3xe^x -3e^x +c}{e^x}
\ ~ \ 
y  = 3x - 3 + \frac{c}{e^x}

$$

perl · Answer

Your work looks good. As Nnesha pointed out, just take the derivative of x. :)