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

2 Answers
Feb 14, 2018

#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)#

Feb 14, 2018

#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)#