Francisco Proctor

2022-07-25

Solve the following initial value problem:

$(1-{t}^{2}){y}^{t}-4ty=1,y(2)=-1$

What is the largest interval on which its solution is guaranteed to uniquely exist?

$(1-{t}^{2}){y}^{t}-4ty=1,y(2)=-1$

What is the largest interval on which its solution is guaranteed to uniquely exist?

neobuzdanio

Beginner2022-07-26Added 13 answers

The homogeneous differential equation can be solved by separation of variables.

$dy/y=4tdt/(1-{t}^{2})$ is the separated differential equation.

Integration $\mathrm{ln}(y)=-2\mathrm{ln}(-1+{t}^{2})$. Exponentiation: $y(t)=a({t}^{2}-1{)}^{-2}.y(2)=a/9=-1.a=-9.y(t)=-9/({t}^{2}-1{)}^{2}$.

The inhomogeneous differential equation is solved by:$y(t)=(-25+3t-{t}^{2})/(3(-1+{t}^{2}{)}^{2})$.This is derived by variation of constants.$y(t)=f(t)(-1+{t}^{2}{)}^{-2}.{y}^{\prime}(t)={f}^{\prime}(t)(-1+{t}^{2}{)}^{2}-4tf(t)(-1+{t}^{2}{)}^{-3}$.So in the inhomogeneous differential equation: $-{f}^{\prime}(t)(-1+{t}^{2}{)}^{-1}+4tf(t)(-1+{t}^{2}{)}^{-2}-4tf(t)(-1+{t}^{2}{)}^{-2}={f}^{\prime}(t)(-1+{t}^{2}{)}^{-1}=1$. Simply integrate and adopt the constant to rewrite the above result.

$dy/y=4tdt/(1-{t}^{2})$ is the separated differential equation.

Integration $\mathrm{ln}(y)=-2\mathrm{ln}(-1+{t}^{2})$. Exponentiation: $y(t)=a({t}^{2}-1{)}^{-2}.y(2)=a/9=-1.a=-9.y(t)=-9/({t}^{2}-1{)}^{2}$.

The inhomogeneous differential equation is solved by:$y(t)=(-25+3t-{t}^{2})/(3(-1+{t}^{2}{)}^{2})$.This is derived by variation of constants.$y(t)=f(t)(-1+{t}^{2}{)}^{-2}.{y}^{\prime}(t)={f}^{\prime}(t)(-1+{t}^{2}{)}^{2}-4tf(t)(-1+{t}^{2}{)}^{-3}$.So in the inhomogeneous differential equation: $-{f}^{\prime}(t)(-1+{t}^{2}{)}^{-1}+4tf(t)(-1+{t}^{2}{)}^{-2}-4tf(t)(-1+{t}^{2}{)}^{-2}={f}^{\prime}(t)(-1+{t}^{2}{)}^{-1}=1$. Simply integrate and adopt the constant to rewrite the above result.

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