Question

Solve the given initial-value problem and give

the largest interval I on which the solution is defined.

$\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{8}}$

Short answer

The largest interval where the solution is defined $y=\frac{1}{{(t+1)}^2-9}\mathrm{where}-4<t<2$

Step-by-step solution

Convert the given differential equation into variable separable form

The given differential equation, can be written as

$\begin{array}{rcl}\int \frac{dy}{y^2}& =& \int -2(t+1)dt\\ -\frac{1}{y}& =& -(t+1{)}^2+c\\ y& =& \frac{1}{{(t+1)}^2+c}y=\frac{1}{{(t+1)}^2+c}\end{array}$

The initial conditions $y\left(0\right)=-\frac{1}{8}$implies

$-\frac{1}{8}=\frac{1}{{(0+1)}^2+c}\Rightarrow c_1=-9$

Solution of the initial value problem is

or$y=\frac{1}{(t^2+2t-8)}$ where $-4<t<2$