How to find integrating factor of ode y’+y=e^(-x) then find solution satisfying y(0)=1 ?

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Introduction : In this question, we will find integrating factor of the ode y’+y=e^(-x) then finding it’s general solution.

Question : Find integrating factor method of the ode

\[y^\prime+y=e^{-x}\]

then find solution satisfying \(y(0)=1\)

Solution : Since given ode

\[y^\prime+y=e^{-x}\]

is a linear ode as it can be written in the form

\[y^\prime+p(x)y=q(x)\]

where \(p(x)=1, q(x)=e^{-x}\)

Now integrating factor (IF) of a Linear ode is obtained as

\[IF=e^{\int{p(x)dx}}\]

So

\[IF = e^{\int{1 dx}} =e^x\]

Now general solution is obtained as

\[y(x)IF =\int q(x)(IF) dx +c\]

\[y(x)e^x =\int e^{-x}e^x dx +c=\int 1 dx +c =x+c \]

multiplying by \(e^{-x}\) both sides

\[y(x)=ce^{-x}+xe^{-x}\]

is the general solution of the ode.

now apply initial condition \(y(0)=1\)

\[1=c+0\]

\[c =1\]

Thus

\[y(x)=e^{-x}+xe^{-x}\]

\[y(x)=e^{-x}(1+x)\]

is the required solution.


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