# I need help to use the method of variation of parameters to find the solution y(x) of the non- homogeneous differential equation y'' + 4y' - 5y = 8e^3x, with initial conditions y(0) = 5 and...

I need help to use the method of variation of parameters to find the solution y(x) of the non- homogeneous differential equation y'' + 4y' - 5y = 8e^3x, with initial conditions y(0) = 5 and y'(0) = 0

Please post as much steps as possible if you can't get the final answer.

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According to the variations of parameters method, the formula for the particular solution of the equation of the form

`y''+q(x)y' + r(x) y = g(x)`

is

`Y_P = -y_1 int y_2(x)g(x)/(W(y_1, y_2)) dx + y_2 int y_1(x)g(x)/(W(y_1, y_2)) dx`

Here,

`y_1` and `y_2`

are the solutions of the corresponding homogeneous equation and `W(y_1, y_2)`

is the Wronskian.

(Please see the reference link for the derivation and the discussion of this formula.)

The given equation is

`y'' + 4y' - 5y = 8e^(3x)`

To use the method of variations, we need to first find the solutions of the corresponding homogeneous equation

`y" + 4y' - 5y = 0`

This equation has the characteristic equation `lambda^2 + 4lambda - 5 = 0` , with the roots `lambda_1 = -5` and `lambda_2 = 1`

` `This means its solution is

`y(x) = c_1e^(-5x) +c_2e^x` , so

`y_ 1 = e^(-5x) ` and `y_2 = e^x` .

The Wronskian of these two functions equals

`y_1y_2' - y_1' y_2 = e^(-5x)e^x - (-5e^(-5x))e^x = e^(-4x) + 5e^(-4x) = 6e^(-4x)`

Now we can proceed to find the particular solution using the formula for the method for the variation of parameters. To make writing easier, I will calculate the two integrals first and then put them back in the formula.

The first integral is

`int y_2(x)g(x)/(W(y_1, y_2))dx = int (e^x*8*e^(3x))/(6e^(-4x))dx = 4/3int e^(8x)dx = 1/6e^(8x)`

The second integral is

`int y_1(x) g(x)/(W(y_1, y_2)) dx = int (e^(-5x)*8*e^(3x))/(6e^(-4x)) dx = 4/3int e^(2x)dx = 2/3e^(2x)`

So, putting these results into the formula for the particular solution, we obtain

`Y_p = -e^(-5x)*1/6*e^(8x) + e^x * 2/3e^(2x) = -1/6e^(3x) + 2/3e^(3x) = =1/2e^(3x)`

We have found the particular solution. The general solution of the non-homogeneous equation is the general solution of the homogeneous equation, which we already have, plus the particular solution of the non-homogeneous equation.

Thus, the general solution is

`y(x) = c_1e^(-5x) + c_2e^x + 1/2e^(3x)`

For the given initial conditions, we can find the constants:

y(0) = 5, so `y(0) = c_1 + c_2 + 1/2 = 5`

y'(0) = 0, so ` `

`y'(0) = -5c_1 + c_2 + 3/2 = 0`

Solving this system of equations results in the values of the constants equal 1 and 3.5.

**So, the solution of the given equation satisfying the given initial conditions is**

`y(x) = e^(-5x) + 7/2e^x + 1/2e^(3x)`

`y'(0) = -5c_1+c_2 + 3/2 = 0`

`romeF`

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