Differential Equation: Check my work please

[vipertongn]

I just want someone to check my work this is the problem

(3x^2 − y)dx + (y − x)dy = 0

I'm going to use the exact equation method to solve it because it seems like the most practical method for the case...I was thinking substitution but the power is not the same for the highest one.

M=3x^2 − y
N =y − x

dM/dy= -1 = dN/dx

fx = 3x^2 - y dx
f = $$\int(3x^2 - y)dx$$ = x^3 - xy +h(y)
fy =S (y-x)dy = y^2/2 -xy + g(x)
h(y)= x^3+c
g(x)= y^2/2+c

Implicit solution is

x^3-xy+y^2/2 +c = 0 or xy-x^3-y^2/2 = c

 

[vela]

Looks good. You can always check your answer by differentiating it and seeing if you get what you started with.

 

[vipertongn]

Thanks, now I'm trying to figure out a new problem (didn't want to make another post)
y'+y/2=x/2y.

I'm trying to figure out what would be a good method to solve this. Any suggestions? I ruled out exact. It looks like a linear equation kinda...but I'm not sure

 

[vela]

Hint: Consider (y²)'.

 

[vipertongn]

wel i know i could have dy/dx = x-y^2/2y But I still don't know which method to solve this. if i try substitution it doesn't work...

Ok I tried this I substituted

u=y^2
du=2ydy

2ydy=(x-y^2)dx
du=(x-u)dx
du/dx+u=x
Makes a Linear Eq
p(x)=1
f(x)=x

e^integral(1)dx = e^x

e^x(u)=integral(e^x(x))dx
e^x(u)=e^x+e^x(x)+c
u=1+x+ce^-x
y^2=1+x+ce^-x
y=sqrt(1+x+ce^-x)

:D right?

 

[vela]

Almost. It doesn't quite work when you substitute it back in the original equation.