What is a solution to the differential equation #dy/dx=xy^2# with the particular solution #y(2)=-2/5#?

Question

Emilia Aguilar · Accepted Answer

# y = - 2/(x^2 + 1)#

#dy/dx=xy^2#

this is separable

#1/y^2 \ dy/dx=x#

#int \ 1/y^2 \ dy/dx \ dx =int \ x \ dx#

#int \ 1/y^2 \ dy =int \ x \ dx#

power rule

# - 1/y = x^2/2 + C#

#y(2) = - 2/5#

# 5/2 = 2 + C implies C = 1/2#

# - 1/y = 1/2 (x^2 + 1)#

# - 2/y = (x^2 + 1)#

# - y/2 = 1/(x^2 + 1)#

# y = - 2/(x^2 + 1)#

Aurora Alvarado · Answer

undefined To find the solution to the differential equation $\frac{dy}{dx} = xy^2$ with the particular solution $y(2) = -\frac{2}{5}$, you can separate variables and integrate both sides.

Separating variables:
$$
\frac{dy}{y^2} = x \, dx
$$

Integrate both sides:
$$
\int \frac{dy}{y^2} = \int x \, dx
$$

For the left integral:
$$
\int \frac{dy}{y^2} = -\frac{1}{y} + C_1
$$

For the right integral:
$$
\int x \, dx = \frac{1}{2}x^2 + C_2
$$

Combine the integrals:
$$
-\frac{1}{y} = \frac{1}{2}x^2 + C
$$
Where $C = C_2 - C_1$.

Given the initial condition $y(2) = -\frac{2}{5}$:
$$
-\frac{1}{-\frac{2}{5}} = \frac{1}{2}(2)^2 + C
$$
$$
-5 = 2 + C
$$
$$
C = -7
$$

The solution is:
$$
-\frac{1}{y} = \frac{1}{2}x^2 - 7
$$

Finally, solve for $y$:
$$
y = -\frac{1}{\frac{1}{2}x^2 - 7}
$$