Differential/Integration equation manipulation

[orangesun]

Homework Statement

a) Solve the differential equation
dy/dx = x(y²+3)/y

b) Find the unique function y(x) satisfying the differential equation with initial
condition
dy/dx = x²y, y(1) = 1

Homework Equations

The Attempt at a Solution

With question a) I am no entirely sure but I have done
dy/dx = x(y²+3)/y
Let u = y²+3/y
So dy/dx = x.u
Integral of (dy/dx) = Integral (x.u)
y = x²u²/2
Then sub it back in. I'm not entirely sure


With question b) I have done
dy/dx = x²y
dy . 1/y = x²dx
integral of(1/y dx) = x³/3 + c
Ln y = x³ /3 + c

I am not sure if I am on the right track,
I would appreciate any help.
Thanks

 

[rock.freak667]

For a), try putting all the 'y's by the dy and all the x's by the dx like you did in part b

Part b is correct. Now just sub in the condition y(1)=1

 

[lanedance]

as rock points out, they're both separable

 

[HallsofIvy]

Homework Statement

a) Solve the differential equation
dy/dx = x(y²+3)/y

This is equivalent to
\frac{y}{y^2+ 3} dy= x dx
Integrate both sides.

b) Find the unique function y(x) satisfying the differential equation with initial
condition
dy/dx = x²y, y(1) = 1

Homework Equations

The Attempt at a Solution

With question a) I am no entirely sure but I have done
dy/dx = x(y²+3)/y
Let u = y²+3/y
So dy/dx = x.u
Integral of (dy/dx) = Integral (x.u)
y = x²u²/2
Then sub it back in. I'm not entirely sure

No, \int (dy/dx) dx= \int x u dx but u is a function of y which is a function of x itself so you really have \int x f(x)dx which cannot be integrated since you don't know "f(x)". Instead separate, as I suggested, and integrate the function of y with respect to y, the function of x with respect to x.

With question b) I have done
dy/dx = x²y
dy . 1/y = x²dx
integral of(1/y dx) = x³/3 + c
Ln y = x³ /3 + c

I am not sure if I am on the right track,
I would appreciate any help.
Thanks

Yes, what you have done for this is correct. Now, use the condition that when x= 1, y= 1 to solve ln(1)= 0= 1³/2+ c for c.