Solve the differential equation: y'= sqrtx * y

Expert Answers
hala718 eNotes educator| Certified Educator

y' = sqrtx * y

To solve differential equation, first we will rewrite the equation:

we know that y' = dy/dx

==> dy / dx = sqrtx * y

Now we will group x terms on one side and y terms on the other side of the equality:

We will multiply by dx/y for both sides:

==> dy/y = sqrtx dx

Now let us integrate both sides:

intg dy/y = intg sqrtx dx

==> ln y = (x^3/2 )/(3/2) + C

==> ln y =( 2/3)(x^3/2) + C

==> y= e^[(2/3)*x^3/2)+ C]

==> y = e^(2/3)x^3/2  + C

neela | Student

y' = sqrtx*y


dy/dx = (sqrtx)*y.

We wrte this  as below:

(1/y) dy = (sqrtx)dx

Now we integrate both sides:

 ln (y) = {x(1/2+1)}/(1/2+1} +C.

ln y =  (2/3)x^(3/2) + C, where c is a constant of integration.

y =  K*e^ ((2/3)*x^(3/2)) , where k is a constant.