Find 2nd order linear homogeneous ODE for which both y1(x)=3e^(2x) - 7e^x . y2(x)=e^(2x) + e^x are solns. Find soln y3(x) such that y3(0)=1, y'3(0)=0.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`y1(x) = 3e^(2x)-7e^x`

Differentiating this would get,

`y'1(x) = 6e^(2x)-7e^x`

Differentitating again,

`y''1(x) = 12e^(2x) -7e^x`

`(d^2y1)/(dx^2) = 12e^(2x) -7e^x`

Following the same procedure for y2(x),

`y2(x) = e^(2x)+e^x`

Differentiating this would get,

`y'2(x) = 2e^(2x)+e^x`

Differentitating again,

`y''2(x) = 4e^(2x)+e^x`

`(d^2y2)/(dx^2) = 4e^(2x)+e^x`

Both these second order ODEs are of the form of,

`(d^2y)/(dx^2) = ae^(2x)+be^x`

Therefore the required solution is,

`(d^2y)/(dx^2) = ae^(2x)+be^x`

To find y3(x) we can integrate the above equations and find values for a and b.

Integrating wrt x,

`(dy)/(dx) = 1/2ae^(2x)+be^x`

But we know, y'3(0) = 0

Therefore,

`0= 1/2a+b`

a+2b = 0 ----1

Integrating again,

`y = 1/4ae^(2x)+be^x`

But y3(0) = 1

So,

`1= 1/4a+b`

a+4b = 4 ----2

solving equationf 1 and equation 2 will give,

a = -4 and b = 2

Therefore y3(x) is given by,

`y3(x) = 1/4(-4)e^(2x)+2e^x`

`y3(x) = -e^(2x)+2e^x`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial