#dy/dx = sqrt(x/y)# ?

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Answer 1 · 2018-04-26T22:08:20

#y=pmsqrt(4/3x^(3//2)+C)#

We have the differential equation:

#dy/dx=sqrtx/y#

Use separation of variables to get #x# and #y# on opposite sides of the equation:

#ycolor(white).dy=sqrtxcolor(white).dx#

Now integrate both sides:

#intycolor(white).dy=intx^(1//2)color(white).dx#

#1/2y^2=1/(3//2)x^(3//2)+C#

#1/2y^2=2/3x^(3//2)+C#

#y^2=4/3x^(3//2)+C#

(I'll just keep writing #C# to represent any arbitrary constant, it doesn't matter that we've multiplied it by two.)

#y=pmsqrt(4/3x^(3//2)+C)#

Answer 2 · 2018-04-27T00:58:25

# y^(3/2) = x^(3/2) + C #

We have:

# dy/dx = sqrt(x/y) #

We can collect terms:

# dy/dx = sqrt(x)/sqrt(y) #

# sqrt(y) \ dy/dx = sqrt(x) #

Which is a Separable DE, so we can "separate the variables" to get

# int \ sqrt(y) \ dy = int \ sqrt(x) \ dx #

Now we can integrate:

# (y^(3/2))/(3/2) = (x^(3/2))/(3/2) + c #

Leading to the General Solution:

# y^(3/2) = x^(3/2) + C #