`y' = sqrt(x)y` Solve the differential equation

Expert Answers
marizi eNotes educator| Certified Educator

An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows `(dy)/(dx)= f(x,y)` .

It can also be in a form of `N(y) dy= M(x) dx` as variable separable differential equation..

 To be able to set-up the problem as `N(y) dy= M(x) dx` , we let` y' = (dy)/(dx).`

 The problem: `y'=sqrt(x)y` becomes:


Rearrange by cross-multiplication, we get:


Apply direct integration on both sides:` int (dy)/y=int sqrt(x)dx`  to solve for the general solution of a differential equation.


For the left side, we applying basic integration formula for logarithm:

`int(dy)/y= ln|y|`

For the right side, we apply the Law of Exponent: `sqrt(x)=x^(1/2)` then follow the Power Rule of integration: `int x^n dx = x^(n+1)/(n+1)+C`

`int sqrt(x)* dx= int x^(1/2)* dx`

                   `= x^(1/2+1)/(1/2+1)+C`

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

                 ` = x^(3/2)*(2/3)+C`

                 ` = (2x^(3/2))/3+C`

Combining the results from both sides, we get the general solution of the differential equation as:



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