solve the ODE y4y4y2ex x y0 2 y0 4Solutiony 4y 4y 2ex x

solve the ODE y\'\'+4y\'+4y=2e^-x +x y(0)= 2 , y\'(0)= -4

Solution

y\" + 4y\' +4y = 2e^-x + x

The characteristic equation corresponding to this is given by:

r²+4r+4 = 0

=> (r+2)² = 0 => r=-2, -2

So the complimentary solution is given by: y_c = Ae^-2x + Bxe^-2x

y(0) = 2 => x=0 then y=2

So 2= A+0 => A=2

Also y\'(0) = -4 => x=0 the y\'=0

So y\'_c = -2Ae^-2x + Be^-2x-2Bxe^-2x

-4 = -4+B-0 => B=0

So y_c = 2e^-2x

Now we find particular solution by assuming, y= Ae^-x +Bx + C

y\' = -Ae^-x+B and y\" = Ae^-x

Substitute these values of y\', y\" and y in original equation and compute for A, B and C , we get:

Ae^-x-4Ae^-x+4B+4Ae^-x+4Bx+4C = 2e^-x+x

Equate coefficients of e^-x on both sides, we get:

A-4A+4A=2 => A=2

Now equating coefficients of x on both sides:

4B = 1 => B=1/4

Equating constants both sides:

4B + 4C = 0 => 4(1/4)+4C = 0 => 1+4C=0 => C=-1/4

So the particular solution is given by:

y_p = 2e^-x+x/4-1/4

Hence the general solution is given by:

y = y_c+y_p = 2e^-2x+2e^-x+x/4-1/4