Math ODE question please help?

Question

How do i solve the following ODEs, tried just about everything i know

a) x^2y' + y^2 = xyy'

It is a first order ODE, the methods i have learnt so far are substitution, exact ODEs, separable ODEs and using integrating factors.

b) 2e^x + dy/dx (1 - e^x)tan y = 0

thanks

Answer 1

a) x^2y' + y^2 = xyy'

y' (xy - x^2) = y^2

y' = y^2/[xy - x^2]

Since x=kx and y = ky does not change anything, try z = y/x

dy/dx = x(dz/dx) + z

y' = y^2/[xy - x^2] = (zx)^2/[x(zx) - (x)^2] = z^2/[z - 1] = z^2/(z - 1)

z^2/(z - 1) = x(dz/dx) + z

x(dz/dx) = [z^2 - z(z - 1)]/(z - 1) = z/(z - 1)

[(z - 1)/z]dz = dx/x

z - ln(z) = ln(x)

(y/x) - ln(y/x) = ln(x)

(y/x) - ln(y) + ln(x) = ln(x)

ln(y) = y/x + C

Check by integrating:

(1/y)y' = (1/x)y' - y/x^2

x^2y' = xyy' - y^2

x^2y' + y^2 = xyy' ... the original equation so the solution is good

b. 2e^x + dy/dx (1 - e^x)tan y = 0 ... just divide by (1 - e^x)

(2e^x)dx/(1 - e^x) + tan(y)dy = 0

for the first term use w = 1 - e^x and dw = -e^x dx

-2ln(1 - e^x) + ln[sec(y)] + C

Check:

2e^x/(1 - e^x) + (1/sec(y)]sec(y)tan(y)(dy/dx) and this looks good