How do you solve #y'=-xy+sqrty# given y(0)=1?

Question

Aurora Alvarado · Accepted Answer

See below.

Making #y = z^2# we obtain #z (x z + 2 z'-1) = 0# or #z=0# and #x z + 2 z'-1=0# We discard #z=0# due to the initial conditions so we follow with #2 z'+x z-1=0# which is a linear non-homogeneous differential equation with solution #z = C e^(-(x^2/4))+ e^(-(x^2/4)) int_0^(x/2)e^(-xi^2) d xi# and then #y = pm sqrt( e^(-(x^2/4))(C+ int_0^(x/2)e^(-xi^2) d xi))# and #y(0) = pm sqrt(C) = 1# then #C = 1# and #y = pm sqrt( e^(-(x^2/4))(1+ int_0^(x/2)e^(-xi^2) d xi))#

Emilia Aguilar · Answer

undefined To solve the initial value problem $ y' = -xy + \sqrt{y} $ with the initial condition $ y(0) = 1 $, we can follow these steps:

1. Separate variables by moving terms involving $ y $ to one side and terms involving $ x $ to the other side:
   $ \frac{dy}{\sqrt{y}} = -xdx $.

2. Integrate both sides:
   $ \int \frac{1}{\sqrt{y}} dy = \int -x dx $.

3. The integral of $ \frac{1}{\sqrt{y}} $ is $ 2\sqrt{y} $, and the integral of $ -x $ is $ -\frac{x^2}{2} $, so we have:
   $ 2\sqrt{y} = -\frac{x^2}{2} + C $, where $ C $ is the constant of integration.

4. Now, apply the initial condition $ y(0) = 1 $:
   $ 2\sqrt{1} = -\frac{0^2}{2} + C $,
   $ 2 = C $.

5. Substitute $ C = 2 $ back into the equation:
   $ 2\sqrt{y} = -\frac{x^2}{2} + 2 $.

6. Solve for $ y $:
   $ \sqrt{y} = -\frac{x^2}{4} + 1 $.

7. Square both sides to solve for $ y $:
   $ y = \left(-\frac{x^2}{4} + 1\right)^2 $.

So, the solution to the initial value problem $ y' = -xy + \sqrt{y} $ with $ y(0) = 1 $ is $ y = \left(-\frac{x^2}{4} + 1\right)^2 $.