If $I$ is an interval of $\mathbb { R }$, we say that $u \in \mathcal { C } ^ { 1 } ( I , \mathbb { R } )$ satisfies (II.1) on $I$ if and only if $$\forall t \in I , \quad u ( t ) \left( u ( t ) + 2 t u ^ { \prime } ( t ) \right) = - 1$$ By stating precisely the theorem used, show that if $( t _ { 0 } , u _ { 0 } )$ is in $\left( \mathbb { R } ^ { * } \right) ^ { 2 }$, there exist an open interval $I$ of $\mathbb { R }$ containing $t _ { 0 }$ and a function $u \in \mathcal { C } ^ { 1 } ( I , \mathbb { R } )$ such that $u$ is a solution of (II.1) on $I$ and satisfies $u \left( t _ { 0 } \right) = u _ { 0 }$.
If $I$ is an interval of $\mathbb { R }$, we say that $u \in \mathcal { C } ^ { 1 } ( I , \mathbb { R } )$ satisfies (II.1) on $I$ if and only if
$$\forall t \in I , \quad u ( t ) \left( u ( t ) + 2 t u ^ { \prime } ( t ) \right) = - 1$$
By stating precisely the theorem used, show that if $( t _ { 0 } , u _ { 0 } )$ is in $\left( \mathbb { R } ^ { * } \right) ^ { 2 }$, there exist an open interval $I$ of $\mathbb { R }$ containing $t _ { 0 }$ and a function $u \in \mathcal { C } ^ { 1 } ( I , \mathbb { R } )$ such that $u$ is a solution of (II.1) on $I$ and satisfies $u \left( t _ { 0 } \right) = u _ { 0 }$.