Question

Solve the initial value first order linear differential equation: y'=2y+x & y(0)=1

Answer 1

@TuringTest

Answer 2

@mahmit2012

Answer 3

\[y'-2y=x\]
\[y_c'-2y_c=0;\ and\ y_p'-2y_p=x\]
\[m=2\ so\ y_c=Ae^{2x}\]
\[Let\ y_p=C_1x+C_2\]
\[Then\ y_p'=C_1\]
\[And\ C_1-2(C_1x+C_2)=x\]
\[So\ C_1-2C_2=0\ \ and \ -2C_1=1\]

Answer 4

Follow me?
\[y=y_c+y_p\]

Answer 5

Wouldn't it be Ae^-2x? Also, how do you find m?

Answer 6

No. If it were Ae^(-2x) then
\[-2Ae^{-2x}-2Ae^{-2x}=0\]
As for m:
\[y_c'−2y_c=0\]
\[m=\frac{y_c'}{y_c}\]
\[m-2=0\]

Answer 7

So would it be:
\[y=-x/2-1/4+(3/4)e ^{-2x}\]

Answer 8

Almost. I think A would need to be 5/4 for y(0) to be 1.

Answer 9

Thanks

Answer 10

And again \[Ae^{2x}\ not\ Ae^{-2x}\]

Answer 11

Got it

Answer 12

Sweet.