Question

How to solve for the differential equation? `sqrt(2xy)dy/dx=1`

Answer 1

`y=3^(2/3)/2(2sqrt(x)+C)^(2/3)`

Explanation:

We want to solve the differential equation

`sqrt(2xy)dy/dx=1`

Use seperation of variable

Hence, move the y terms to one side of the equation,
and move the x terms to the other side

`sqrt(2)sqrt(y)dy=1/sqrt(x)dx`

Integrate both sides by the power rule

`sqrt(2)intsqrt(y)dy=int1/sqrt(x)dx`

`2/3sqrt(2)y^(3/2)=2sqrt(x)+C`

Solve for `y`

`2^(3/2)y^(3/2)=3(2sqrt(x)+C)`

`y^(3/2)=2^(-3/2)*3(2sqrt(x)+C)`

`y=3^(2/3)/2(2sqrt(x)+C)^(2/3)`

Answer 2

`y=((3sqrt2)/2x^(1/2)+c_2)^(2/3)`

Explanation:

`sqrt(2xy)(dy)/(dx)=1`

separate variables

`sqrt2intsqrtydy=intdx/sqrtx`

changing to index form

`sqrt2inty^(1/2)dy=intx^(-1/2)dx`

using the power rule for integration

`color(blue)(intx^ndx=x^(n+1)/(n+1)+c,n!=-1`

we have

`sqrt2(2/3)y^(3/2)=2x^(1/2)+c_1`

`y^(3/2)=3/sqrt2x^(1/2)+c_2`

`y=((3sqrt2)/2x^(1/2)+c_2)^(2/3)`