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Consider the initial value problem ty'-y=t^{2}\sin ty(t_{0})=y_{0}. Show that there is a unique solution when t_{0}\neq 0, no solution when t_{0}=0 and y_{0}\neq 0 and an infinite number of solutions when t_{0}=y_{0}=0. Explain these reults using appropriate existence and uniqueness theorems.

y'-{\frac  {1}{t}}y=t\sin(t)\,

Find the integrating factor:

e^{{-\int {\frac  {1}{t}}dt}}=e^{{\ln {\frac  {1}{t}}}}={\frac  {1}{t}}\,

{\frac  {1}{t}}y'-{\frac  {1}{t^{2}}}y=\sin t\,

{\frac  {d}{dt}}\left({\frac  {1}{t}}y\right)=\sin t\,

{\frac  {1}{t}}y=-\cos t+c_{1}\,

y(t)=-t\cos t+c_{1}t\,

The inital condition gives

y(t_{0})=-t_{0}\cos t_{0}+c_{1}t_{0}=y_{0}\,

c_{1}={\frac  {y_{0}+t_{0}\cos t_{0}}{t_{0}}}={\frac  {y_{0}}{t_{0}}}+\cos t_{0}\,

so

y(t)=-t\cos t+{\frac  {ty_{0}}{t_{0}}}+t\cos t_{0}\,

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