Question

Find general solution of `dy/dx = (x+y)^2/(xy)` ?

Answer 1

Please see the explanation below

Explanation:

The ode is

`dy/dx=(x+y)^2/(xy)`

`dy/dx=(x^2+y^2+2xy)/(xy)`

`dy/dx=x/y+y/x+2`

Let `y=vx`, then

`dy/dx=v+x(dv)/dx`

`v+x(dv)/dx=1/v+v+2`

`x(dv)/dx=1/v+2=(1+2v)/v`

Now, separate the variables

`v/(1+2v)dv=dx/x`

`intv/(1+2v)dv=intdx/x`

Let `1+2v=u`, `=>`, `2dv=du`

`intv/(1+2v)dv=1/2int((u-1)du)/u`

`=1/2int(1-1/u)du`

`=1/2u-1/2lnu`

`=1/2(1+2v)-1/2ln(1+2v)`

Therefore,

`1/2(1+2v)-1/2ln(1+2v)=lnx+C`

`(1+y/x)-ln(1+2y/x)=2lnx+C_1`

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

`1+y/x=lnkx(x+2y)`

`y/x=lnkx(x+2y)-1` where `k in RR`

Answer 2

`2y=xln{(2y+x)x^3c^4}`.

Explanation:

We have, `dy/dx=(x+y)^2/(xy)={x(1+y/x)}^2/(xy)`.

`:. dy/dx=x^2/(xy)(1+y/x)^2`

`i.e., dy/dx=x/y(1+y/x)^2`.

So, this is a Homogeneous Diff. Eqn.

Subst.ing `y/x=v, i.e., y=vx," we have, "dy/dx=x(dv)/dx+v`.

`:. x(dv)/dx+v=1/v(1+v)^2=1/v+2+v`.

`:. x(dv)/dx=1/v+2=(1+2v)/v`.

`:. v/(2v+1)dv=dx/x............"[Separable Variable]"`.

`:. dx/x=1/2*(2v)/(2v+1)dv=1/2{(2v+1)-1}/(2v+1)dv`,

`=1/2{(2v+1)/(2v+1)-1/(2v+1)}dv`.

`rArr dx/x=1/2(1-1/(2v+1))dv`.

`"Integrating, "intdx/x+lnc=1/2int(1-1/(2v+1))dv`.

`:. lnx+lnc=1/2{v-1/2*ln(2v+1)}=1/2v-1/4ln(2v+1)`.

`:. lnx+lnc+1/4ln(2v+1)=1/2v`,

`or, 4lnx+4lnc+ln(2v+1)=2v`.

`:. ln{(2v+1)(cx)^4}=2v`.

Replacing `v" by "y/x`, we get,

`2y=xln{(2y+x)x^3c^4}`, as the desired General Solution!

Enjoy Maths.!