differential equations?

Question

solve the following differential equations
I am having trouble with starting the following equations
a) dy/dx + ysec(x)tan(x) = e^-sec(x) , y(0) = 0

b) dy/x = (x^2 +3y^2)/2xy y(1) = 2
[HINT - on the right side, divide the top and bottom by x^s and use substitution v = y/x]

yes it is already differentiated and the information provided is given

Answer 1

for b, is (x^2 +3y^2)/2xy the equation already differentiated? and is y(1) = 2 given?

Answer 2

a) This is a first order DE. The integrating factor is e^secx. That is, when both sides are multiplied by this factor the left side becomes d/dx of (y e^secx) and the right side is e^(2secx). I don't think e^(2secx) can be integrated in a closed form. So if we let J = the integral of e^(2secx) then an answer is

cosx * [ J + constant].

b) Now use the hint which works for "homogenous DE's" : v = y/x. So vx=y and vdx + xdv = dy. Substitute this in and replace y by xv to get

2(x^2)v(xdv + vdx) = [x^2 + 3(x^2)(v^2)]dx.

Divide by x^2 and rearrange the terms to get:

2xvdv = (1 + v^2)dx.

Separate variables and integrate to get

ln(1+v^2) = ln abs(x) + c.

So 1 + v^2 = K*abs(x). Replace v with y/x to get

y = +/- sqr[x^2 (K*abs(x) - 1)].

Use the initial condition to find K=5 and y= sqr[x^2 (5*abs(x) - 1)].