Can someone walk me through on solving this DE?
The general solution to the differential equation y' = y + x^2 is y =

Question

Can someone walk me through on solving this DE?
The general solution to the differential equation y' = y + x^2 is y =

amistre64 · Answer

get the y parts to one side

bahrom7893 · Answer

integrating factors?

bahrom7893 · Answer

y' - x^2 = y

amistre64 · Answer

y' -y = x^2

then solve the characteristic eq

r-1 = 0 when r = 1

$$y=c_1e^x+yp$$
maybe

amistre64 · Answer

or since its a first order we can go to the reversal of int by parts

rogue · Answer

Ahh, nevermind, I realize my mistake. Thanks anyways =)

bahrom7893 · Answer

wait i moved the wrong thing over lol
y' - y = x^2
e^(Int(-1)) = e^(-x)
mehh lol im too slow haha

amistre64 · Answer

$$y' -y = x^2$$

$$e^{-ln(x)}(y' -y = x^2)$$

$$e^{-ln(x)}y =\int  x\ dx$$

$$y/x =\frac{1}{2}x^2+c$$

$$y =\frac{1}{2}x^3+Cx$$
might be one solution

amistre64 · Answer

the wolf like the first try

rogue · Answer

I got this
$$y' - y = x^2$$$$y' e^{-x} + ye^{-x} = x^2 e^{-x} \rightarrow \frac {d}{dx} \left[ ye^{-x} \right] = x^2 e^{-x}$$$$\int\limits \frac {d}{dx} \left[ ye^{-x} \right] dx = \int\limits  x^2 e^{-x} dx$$$$y e^{-x} = -e^{-x} (x^2 + 2x +2) + C$$$$y = Ce^{x} -x^2 - 2x - 2 $$

amistre64 · Answer

thats it

amistre64 · Answer

lol, I did something odd; shoulda been $\int -1 dx=-x$

amistre64 · Answer

$$yp = ax^2+bx+c$$
$$yp' = 2ax+b$$

$$yp'-yp = 2ax+b-ax^2-bx-c=x^2$$

$$\begin{matrix} (-a)x^2 & (2a-b)x&(b-c)\ 
(1)x^2&0(x)&0\
a=-1&b=-2&c=-2
\end{matrix}$$
$$yp = -x^2-2x-2$$
$$y=c_1e^x-x^2-2x-2 $$

rogue · Answer

Oh, thats interesting! thank you =)