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

User Dodol
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1 Answer

3 votes

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).

User Joe Orost
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