Question

Dy − (y − 5)^2 dx = 0

Answer 1

y=(1-5x+5C)/(C-x)

Explanation:

dy − (y − 5)^2 dx = 0

Add (y-5)^2 dx on both sides:

dy=(y-5)^2 dx

Divide both sides by (y-5)^2:

dy/(y-5)^2=dx

We have separated the variables and are thus ready to integrate:

(y-5)^(-1)/(-1)+C=x

-1/(y-5) + C=x

Perhaps you want to solve for y:

Multiply both sides by (y-5):

-1+C(y-5)=x(y-5)

Subtract C(y-5) on both sides:

-1=x(y-5)-C(y-5)

Distribute:

-1=xy-5x-Cy+5C

Group y terms together:

-1=-5x+5C+xy-Cy

Factor the y out from the terms containing y:

-1=-5x+5C+y(x-C)

Subtract 5C and -5x on both sides:

-1--5x-5C=y(x-C)

Divide both sides by (x-C):

(-1+5x-5C)/(x-C)=y

Multiply by 1=-1/-1:

(1-5x+5C)/(C-x)=y

y=(1-5x+5C)/(C-x)

Answer 2

`(1)/((y-5)) + x +K =0`

Explanation:

In order to solve the first order differential equation given, you have to obtain an equivalent expression with a function of x with dx and a function of y with dy

Ordering the differential equation:

`dy=(y-5)^(2)dx\\`

Multipliying both sides by
`(1)/((y-5)^(2) )`

`(1)/((y-5)^(2) ) dy= dx`

Now, you have to apply the indefinite integral in both sides

`\int\limits {(1)/((y-5)^(2) ) } \, dy=\int\limits \, dx\\ \int\limits {(y-5)^(-2) } \, dy = \int\limits \, dx\\((y-5)^(-2+1) )/(-2+1) + k1 = x + k2\\x+(1)/((y-5))+K=0`

Where K is the algebraic sum of all the constants of integration.

Notice that the integration of a function with the form
`F(y)^(n)` is:

`(F(y)^(n+1) )/(n+1)`