# dy/dx = 6 - y Solve the differential equation

Apply direct integration both sides: intN(y) dy= int M(x) dx to solve for the  general solution of a differential equation.

For the given first order ODE: (dy)/(dx)=6-y  it can be rearrange by cross-multiplication into:

(dy)/(6-y)=dx

Apply direct integration on both sides: int(dy)/(6-y)=int dx

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

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Apply direct integration both sides: intN(y) dy= int M(x) dx to solve for the  general solution of a differential equation.

For the given first order ODE: (dy)/(dx)=6-y  it can be rearrange by cross-multiplication into:

(dy)/(6-y)=dx

Apply direct integration on both sides: int(dy)/(6-y)=int dx

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

u=6-y  then  du = -dy    or   -du=dy

The integral becomes: int(dy)/(6-y)=int(-du)/(u)

Applying basic integration formula for logarithm:

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

Plug-in u = 6-y on   -ln|u| , we get:

int(dy)/(6-y)=-ln|6-y|

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|6-y|= x+C

y =6-e^((-x-C))

or

`y = 6-Ce^(-x)

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