# dy/dx = y + 3 Solve the differential equation

Recall that in solving simple first order "ordinary differential equation" (ODE),  we may apply variable separable differential equation wherein:

N(y)y'=M(x)

N(y)(dy)/(dx)=M(x)

N(y) dy=M(x) dx

Before we can work on the direct integration:  int N(y) dy= int M(x) dx to solve for the  general solution of a differential equation.

For the given...

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Recall that in solving simple first order "ordinary differential equation" (ODE),  we may apply variable separable differential equation wherein:

N(y)y'=M(x)

N(y)(dy)/(dx)=M(x)

N(y) dy=M(x) dx

Before we can work on the direct integration:  int N(y) dy= int M(x) dx to solve for the  general solution of a differential equation.

For the given first order ODE: (dy)/(dx)=y+3  can be rearrange by cross-multiplication into:

(dy)/(y+3)=dx

Apply direct integration on both sides: int(dy)/(y+3)=int  dx

For the left side, we consider u-substitution by letting:

u= y+3 then du = dy

The integral becomes:

int(dy)/(y+3)=int(du)/(u)

Applying basic integration formula for logarithm:

int(du)/(u)= ln|u|

Plug-in u = y+3 on ln|u | , we get:

int(dy)/(y+3)=ln|y+3|

For the right side, we apply the basic integration: int dx= x+C

Combing the results from both sides, we get the general solution of the differential equation as:

ln|y+3|= x+C

or

y =e^(x+C)-3

y =Ce^x-3

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