 # Solve the following initial value problem: (1-t^2)y'-4ty =1, y(2)=-1 What is the largest interval on which its solution is guaranteed to uniquely exist? Francisco Proctor 2022-07-25 Answered
Solve the following initial value problem:
$\left(1-{t}^{2}\right){y}^{t}-4ty=1,y\left(2\right)=-1$
What is the largest interval on which its solution is guaranteed to uniquely exist?
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The homogeneous differential equation can be solved by separation of variables.
$dy/y=4tdt/\left(1-{t}^{2}\right)$ is the separated differential equation.
Integration $\mathrm{ln}\left(y\right)=-2\mathrm{ln}\left(-1+{t}^{2}\right)$. Exponentiation: $y\left(t\right)=a\left({t}^{2}-1{\right)}^{-2}.y\left(2\right)=a/9=-1.a=-9.y\left(t\right)=-9/\left({t}^{2}-1{\right)}^{2}$.
The inhomogeneous differential equation is solved by:$y\left(t\right)=\left(-25+3t-{t}^{2}\right)/\left(3\left(-1+{t}^{2}{\right)}^{2}\right)$.This is derived by variation of constants.$y\left(t\right)=f\left(t\right)\left(-1+{t}^{2}{\right)}^{-2}.{y}^{\prime }\left(t\right)={f}^{\prime }\left(t\right)\left(-1+{t}^{2}{\right)}^{2}-4tf\left(t\right)\left(-1+{t}^{2}{\right)}^{-3}$.So in the inhomogeneous differential equation: $-{f}^{\prime }\left(t\right)\left(-1+{t}^{2}{\right)}^{-1}+4tf\left(t\right)\left(-1+{t}^{2}{\right)}^{-2}-4tf\left(t\right)\left(-1+{t}^{2}{\right)}^{-2}={f}^{\prime }\left(t\right)\left(-1+{t}^{2}{\right)}^{-1}=1$. Simply integrate and adopt the constant to rewrite the above result.