# 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.

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### 1 Answer

`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`