# Solve the following initial value problem for the function y(x). y'=2xy^2;\

Solve the following initial value problem for the function y(x).
You can still ask an expert for help

## Want to know more about Derivatives?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Bertha Jordan
${y}^{\prime }=2x{y}^{2}$
$⇒\frac{dy}{dx}=2x{y}^{2}$
$⇒\int \frac{dy}{{y}^{2}}=\int 2xdx$
$=-{y}^{-1}={x}^{2}+c$
$⇒\frac{-1}{y}={x}^{2}+c$
given $y\left(1\right)=\frac{1}{2}$
$⇒-2=1+c$
$⇒c=-3$
$\therefore \frac{-1}{y}={x}^{2}-3$
$⇒y=\frac{1}{3}-{x}^{2}$ - is the solution
###### Not exactly what you’re looking for?
GaceCoect5v

Solve the separable equation $\frac{dy\left(x\right)}{dx}=2xy{\left(x\right)}^{2}$, such that y(1) = 1/2:
Divide both sides by $y{\left(x\right)}^{2}$:
$\frac{\frac{dy\left(x\right)}{dx}}{y}{\left(x\right)}^{2}=2x$
Integrate both sides with respect to $x$:
$\int \frac{\frac{dy\left(x\right)}{dx}}{y}{\left(x\right)}^{2}dx=\int 2xdx$
Evaluate the integrals:
$-\frac{1}{y}\left(x\right)={x}^{2}+{c}_{1}$, where ${c}_{1}$ is an arbitrary constant.
Solve for $y\left(x\right)$:
$y\left(x\right)=-\frac{1}{{x}^{2}+{c}_{1}}$
Solve for ${c}_{1}$ using the initial conditions:
Substitute $y\left(1\right)=\frac{1}{2}\to y\left(x\right)=-\frac{1}{{x}^{2}+{c}_{1}}$:
$-\frac{1}{{c}_{1}+1}=\frac{1}{2}$
Solve the equation:
${c}_{1}=-3$
Substitute ${c}_{1}=-3\to y\left(x\right)=-\frac{1}{{x}^{2}+{c}_{1}}$:
$y\left(x\right)=\frac{1}{-{x}^{2}+3}$

user_27qwe