Question

solve the initial value problem

Answer 1

\[y \prime =\frac{ x^2+3x+2 }{ y-2 }\]
\[y(1)=4\]

Answer 2

Soooo..... I separated the variables
\[y \prime (y-2)=x^2+3x\]

Answer 3

+2*

Answer 4

and then I went to integrate them... which got me here
\[\frac{ -y^2 }{ 2 }-2y+c=\frac{ x^3 }{ 3 }+\frac{ 3x^2 }{ 2 }+2x\]

Answer 5

but I feel like I am wrong

Answer 6

it looks fine to me, but the actual "separation of hte variables" is a step that you skipped. you should have written
\[y'(y-2)=x^2+3x+2\\\frac{dy}{dx}(y-2)=x^2+3x+2\\(y-2)dy=(x^2+3x+2)dx\\integrate...\]

Answer 7

the separation of the variables is the part where you move the dx and dy on different sides, not the variables themselves
your answer is correct though, except for the +C from the constant

Answer 8

Ooooh. Okay well what do you mean about the +c? Does it go on both sides? Lol I just always add it in

Answer 9

well it technically goes on both sides so you would get
\[f(y)+C_1=f(x)+C_2\]but since they are constants we can just say\[f(y)=f(x)+(C_2-C_1)=f(x)+C_3\] since subtracting or adding two constants we don't know the value of just makes nother constant we also don't know the value of

Answer 10

but you *always* get a +C when you do an indefinite integral

Answer 11

So, when I solve for the initial value problem I'm solving for say C3 on the right hand side?

Answer 12

correct :)

Answer 13

okay, thank you :)

Answer 14

welcome!