# Differential/Integration equation manipulation

1. May 12, 2010

### orangesun

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

2. May 12, 2010

### 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

3. May 12, 2010

### lanedance

as rock points out, they're both separable

4. May 12, 2010

### HallsofIvy

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

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.

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.