Question #42c59

1 Answer
Dec 4, 2017

y=1/2e^(2x+C)+2

Explanation:

.

d/dx(y)=2y-4

First Order Separable Ordinary Differential Equation has the form of:

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

N(y)y'=M(x)

N(y)=1/(2y-4)

M(x)=1

int1/(2y-4)dy=int1dx

1/2ln(2y-4)+C_2=x+C_1

Now, we combine constants:

1/2ln(2y-4)=x+C

ln(2y-4)=2x+C

Using the definition of Log, we isolate y:

e^(2x+C)=2y-4

2y=e^(2x+C)+4

y=1/2e^(2x+C)+2