Question

Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients. y'' − 5y' + 6y = 2et

Answer 1

Therefore the complete primitive is

`y=c_1 e^(2y)+c_2e^(3t)+e^(t)`

Therefore the general solution is

`y=c_1e^(2t)+c_2e^(3t)+e^t`

Explanation:

Given Differential equation is

`y''-5y'+6y=2e^t`

Method of variation of parameters:

Let
`y=e^(mt)` be a trial solution.

`y'= me^(mt)`

and
`y''= m^2e^(mt)`

Then the auxiliary equation is

`m^2e^(mt)-5me^(mt)+6e^(mt)=0`

`\Rightarrow m^2-5m+6=0`

`\Rightarrow m^2 -3m -2m +6=0`

`\Rightarrow m(m -3) -2(m -3)=0`

`\Rightarrow (m-3)(m-2)=0`

`\Rightarrow m=2,3`

∴The complementary function is
`C_1e^(2t)+C_2e^(3t)`

To find P.I

First we show that
`e^(2t)` and
`e^(3t)` are linearly independent solution.

Let
`y_1=e^(2t)` and
`y_2= e^(3t)`

The Wronskian of
`y_1` and
`y_2` is
`\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|`

`=\left|\begin{array}{cc}e^(2t)&e^(3t)\\2e^(2t)&3e^(3t)\end{array}\right|`

`=e^(2t).3e^(3t)-e^(2t).2e^(3t)`

`=e^(5t)` ≠ 0

∴
`y_1` and
`y_2` are linearly independent.

Let the particular solution is

`y_p=v_1(t)e^(2t)+v_2(t)e^(3t)`

Then,

`Dy_p= 2v_1(t)e^(2t)+v'_1(t)e^(2t)+3v_2(t)e^(3t)+v'_2(t)e^(3t)`

Choose
`v_1(t)` and
`v_2(t)` such that

`v'_1(t)e^(2t)+v'_2(t)e^(3t)=0` .......(1)

So that

`Dy_p= 2v_1(t)e^(2t)+3v_2(t)e^(3t)`

`D^2y_p= 4v_1(t)e^(2t)+9v_2(t)e^(3t)+ 2v'_1(t)e^(2t)+3v'_2(t)e^(3t)`

Now

`4v_1(t)e^(2t)+9v_2(t)e^(3t)+ 2v'_1(t)e^(2t)+3v'_2(t)e^(3t)-5[2v_1(t)e^(2t)+3v_2(t)e^(3t)] +6[v_1e^(2t)+v_2e^(3t)]=2e^t`

`\Rightarrow 2v'_1(t)e^(2t)+3v'_2(t)e^(3t)=2e^t` .......(2)

Solving (1) and (2) we get

`v'_2=2 e^(-2t)` and
`v'_1(t)=-2e^(-t)`

Hence

`v_1(t)=\int (-2e^(-t)) dt=2e^(-t)`

and
`v_2=\int 2e^(-2t)dt =-e^(-2t)`

Therefore
`y_p=(2e^(-t)) e^(2t)-e^(-2t).e^(3t)`

`=2e^t-e^t`

`=e^t`

Therefore the complete primitive is

`y=c_1 e^(2y)+c_2e^(3t)+ e^(t)`

Undermined coefficients:

∴The complementary function is
`C_1e^(2t)+C_2e^(3t)`

The particular solution is
`y_p=Ae^t`

Then,

`Dy_p= Ae^t` and
`D^2y_p=Ae^t`

`\therefore Ae^t-5Ae^t+6Ae^t=2e^t`

`\Rightarrow 2Ae^t=2e^t`

`\Rightarrow A=1`

`\therefore y_p=e^t`

Therefore the general solution is

`y=c_1e^(2t)+c_2e^(3t)+e^t`