# `(2+x)y' = 3y` Find the general solution of the differential equation Recall that `y'` is the same as `(dy)/(dx)` . Then in the given problem: `(2+x)y'=3y` , we may write it as:

`(2+x) (dy)/(dx) = 3y.`

This will help to follow the variable separable differential equation in a form of `N(y) dy = M(x) dx.`

To rearrange `(2+x) (dy)/(dx) = 3y` ,cross-multiply `(dx)`...

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Recall that `y'` is the same as `(dy)/(dx)` . Then in the given problem: `(2+x)y'=3y` , we may write it as:

`(2+x) (dy)/(dx) = 3y.`

This will help to follow the variable separable differential equation in a form of `N(y) dy = M(x) dx.`

To rearrange `(2+x) (dy)/(dx) = 3y` ,cross-multiply `(dx)` to the other side:

`(2+x)dy =3y dx`

Divide both sides by `(2+x)` :

`((2+x)dy)/(2+x) =(3y dx)/(2+x)`

`dy =(3y dx)/(2+x)`

Divide both sides by `y` :

`(dy )/y=(3y dx)/((2+x)y)`

`(dy)/y=(3dx)/(2+x)`

To solve for the general solution of the differential equation, apply direct integration on both sides:

`int (dy)/y=int (3dx)/(2+x)`

For the left side, apply the basic integration formula for logarithm

`int (dy)/y= ln|y|`

For the right side, we may apply the basic integration property: `int c*f(x) dx = c int f(x)dx` .

`int (3dx)/(2+x)= 3 int (dx)/(2+x)`

Let `u =2+x` then du= dx

The integral becomes:

`3 int (dx)/(2+x) = 3 int (du)/u`

We can now apply the  basic integration formula for logarithm on the integral part:

`3 int (du)/u= 3ln|u| +C`

Recall `u =(2+x) ` then `3 int (dx)/(2+x) =3ln|2+x| +C`

Combining the results from both sides, we get:

`ln|y|=3ln|2+x| +C`

`y=e^(3ln|x+2|+C)`

` y= e^(ln(x+2)^3+C)`

Law of Exponents:` x^(n+m)= x^n*x^m`

`y= e^(ln(x+2)^3)*e^C`

`e^C =C` is an arbitrary constant, so

`y= Ce^(ln(x+2)^3)`

` y = C(x+2)^3`

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