Question

Dy − (y − 5)^2 dx = 0

Answer 1

Answer:

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

Step-by-step 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

Answer:

$\frac{1}{(y-5)} + x +K =0$

Step-by-step 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 $\frac{1}{(y-5)^{2} }$

$\frac{1}{(y-5)^{2} } dy= dx$

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

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

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:

$\frac{F(y)^{n+1} }{n+1}$