Question

Suppose that the first order differential equation $\mathbf{dy}/\mathbf{dx}=\mathbf{f}\left(\mathbf{x},\mathbf{y}\right)$possess a one parameter family of solutions and that $\mathbf{f}$left( $\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\backslash }\mathit{r}\mathit{i}\mathit{g}\mathit{h}\mathit{t}\mathbf{)}\mathbf{\backslash }\mathbf{]}$satisfies the hypothesis of theorem 1.2.1 in some rectangular region R of xy-plane. Explain why two different solution curve cannot intersect or tangent to each other at a point $\left(x_{0,}{y}_0\right)$in R.

Short answer

The different integral curves cannot intersect because every point lies on exactly one integral curve, so two different solution curve cannot intersect or tangent to each other.

Step-by-step solution

Definition of first order differential equation

The first order differential equation is in the form $F(t,y,y\text{'})=0$. A solution that makes first order differential equation function f(t) makes$F(t,f(t),f\text{'}(t\left)\right)=0$

Step2: Explanation using theorem

• The theorem is about the exsistence of the unique solution
• The rectangular R of the xy-plane is satisfied by the theorem and it has a unique and single solution at any point in the region R
• Here ,it implies that it cannot be two different solutions through any point R and the two different solution curves cannot intersect at tangent to each other
• And if they intersect it does been a unique solution and also it becomes intial-value problem (IVP)
• Also that the different integral curves cannot intersect because every point lies on exactly one integral curve, so two different solution curve cannot intersect or tangent to each other