# Differential Equations... Help please?

I really need your help coz now I'm currently stucked at Medium Level :(

and I really don't know how to deal on this things..

Solve the given initial-value problem.

11. xy^2(dy/dx) = y^3 - x^3, y(1) = 2

12. (x^2 + 2y^2)dx/dy = xy, y(-1) = 1

13. (x + ye^y/x)dx - xe^y/x dy = 0, y(1) = 0

14. ydx + x(ln x - ln y - 1)dy = 0, y(1) = e

Solve the given differential equation by using an appropriate substitution.

15. x(dy/dx) + y = 1/y^2

16. dy/dx - y = (e^x)(y^2)

17. dy/dx = y(xy^3 - 1)

18. x(dy/dx) - (1 + x)y = xy^2

Thank you in advance for helping Ladies and Gentlemen..

Feel free to tell me some hints coz I really neeeed em right now..

Take care always and God bless!

### 1 Answer

- az_lenderLv 77 years agoFavorite Answer
The first two are like the ones you presented yesterday,

they can be solved with substitution u = y/x, xu' + u = y'.

#11. dy/dx = y^3/xy^2 - x^3/xy^2 = (y/x) + 1/(y/x)

xu' + u = u + 1/u

xu' = 1/u

u du = dx/x

(1/2)u^2 = ln(x) + C

(1/2)(y^2/x^2) = ln(x) + C

Use y(1) = 2:

(1/2)(4/1) = ln(1) + C

2 = C

Solution:

(1/2) (y^2/x^2) = ln(x) + 2

or, y^2 = (x^2) [2 ln(x) + 4]

or, whatever other way you want to write it.

#12 - try the same substitution

#13. After staring at this for a while, I realized that you meant y/x to be the exponent.

You have EGREGIOUSLY mistyped this. You didn't follow my advice yesterday

to consult the "Order of Operations" article in Wikipedia. If you refuse to learn

elementary algebra, you will crash and burn in professional life, if someone is

dumb enough to hire you. OK, rant off...

[ x + ye^(y/x) ] dy - xe^(y/x) dx = 0

dy/dx = x e^(y/x) / [x + ye^(y/x)] = e^(y/x) / [ 1 + (y/x)e^(y/x) ]

Same old substitution...

xu' + u = e^u / [ 1 + ue^u ]

xu' = [e^u - u - u^2 e^u] / [1 + ue^u]

(1 + ue^u) du/(e^u - u - u^2 e^u) = dx/x

Integral of the RHS is obviously ln(x) + C.

Integral of the LHS looks tough, trying Wolfram Integrator if you're not getting anywhere.

#14. no instinct about how to do this

#15. It's separable, I don't see the need for a substitution at first.

x dy/dx = 1/y^2 - y = (1 - y^3) / y^2

y^2 dy/(1 - y^3) = dx/x

NOW let u = 1-y^3, du = -3y^2 dy, and you have

(-1/3) du/u = dx/x

(-1/3) ln|u| = ln|x| + C1

1/u = C2 x^3

C = x^3(1-y^3)

I don't know how to do #16 - #18 but will return to this if I think of something