Question

For ODE y”+5y’+4y=2e^(-2x) what is its particular intergral

Answer 1

We have the differential equation,
`y''+5y'+4y=2e^(-2x)`. The question asks to find the particular solution to the DE.

To find the particular solution look at the right-hand-side and make a “guess” of what the solution could be.

Since we have
`2e^(-2x)` we can guess that the particular solution could be in the form
`Ae^(-2x)`, where
`A` is some unknown constant.

Now finding the value of the unknown constant,
`A`. Lets say
`g(x)=Ae^(-2x)`, find
`g'(x)` and
`g''(x)`.

`g(x)=Ae^(-2x)`

=>
`g'(x)=-2Ae^(-2x)`

=>
`g''(x)=4Ae^(-2x)`

Now plug
`g(x)`,
`g'(x)\\`, and
`g''(x)` into the given differential equation.

=>
`y''+5y'+4y=2e^(-2x)`

=>
`(4Ae^(-2x))+5(-2Ae^(-2x))+4(Ae^(-2x))=2e^(-2x)`

=>
`4Ae^(-2x)-10Ae^(-2x)+4Ae^(-2x)=2e^(-2x)`

=>
`8Ae^(-2x)-10Ae^(-2x)=2e^(-2x)`

=>
`-2Ae^(-2x)=2e^(-2x)`

Compare the coefficients,

=>
`-2A=2`

=>
`A=-1`

So we can say our "guess" to the particular solution of the given differential equation is,
`y_(p) =-e^(-2x)`.