`y' + 3y = e^(3x)` Solve the first-order differential equation

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Given` y'+3y=e^(3x)`

when the first order linear ordinary differential equation has the form of


then the general solution is ,

`y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)`




on comparing both we get,

`p(x) = 3 and q(x)=e^(3x)`

so on solving with the above general solution we get:

y(x)=`((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)`

=`((int e^(int 3 dx) *(e^(3x))) dx +c)/e^(int 3 dx)`

first we shall solve

`e^(int 3 dx)=e^(3x) `     


proceeding further, we get

y(x) =`((int e^(int 3 dx) *(e^(3x))) dx +c)/e^(int 3 dx)`

=`((int e^(3x) *(e^(3x))) dx +c)/e^(3x)`

=`((int e^(6x) ) dx +c)/e^(3x)`

= `(e^(6x)/6 +c)/e^(3x)`

=`(e^(6x)/6 +c)*e^(-3x)`

so `y(x)=(e^(6x)/6 +c)*e^(-3x)`

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