Need help with this guys....

Solve the differential equation dy/dx = 4xy^2 and write the general solution in the form y^r + kx^n = C where r, k and n are numbers and C is the arbitary constant.

Thanks...

Printable View

- February 12th 2009, 02:13 AMmr_motivatorDifferential Equation
Need help with this guys....

Solve the differential equation dy/dx = 4xy^2 and write the general solution in the form y^r + kx^n = C where r, k and n are numbers and C is the arbitary constant.

Thanks... - February 12th 2009, 02:27 AMmcmaster
(1/y^2)dy=4xdx

integral of ((1/y^2)dy)= integral of (4xdx)

-(1/y)+C1=2x^2+C2

y= -(1/(2x^2))+C assume C2-C1=C

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

so r=1

k=1/2

n=-2 - February 12th 2009, 12:43 PMMuzza8888
In this question maple has given me a different set of variables that doesnt work out so nicely.

dy/dx = -5*x*y^4

int(y^-4)dy = int (-5*x) dx

1/3y^-3 - 5/2x^2 = C

So when asked for the form y^r + kx^n = C, i have a 2nd constant before the y, namely 1/3, that is damaging to the answer. How could i have done this euation differently to avoid this error?