`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
`y'+p(x)y=q(x)`
then the general solution is ,
`y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)`
so,
`y'+3y=e^(3x)--------(1)`
`y'+p(x)y=q(x)---------(2)`
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) `
so
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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