91Ó°ÊÓ

Suppose that y(t) solves \({\bf{y''(t) = f(y)}}\) where f is a continuous function independent of y’

Or t. Suppose that F(y) is an indefinite integral of f(y) and \({\bf{f(y) = }}\frac{{\bf{d}}}{{{\bf{dy}}}}{\bf{(F(y)}}\)then\(\frac{{\bf{1}}}{{\bf{2}}}{{\bf{(y')}}^{\bf{2}}}{\bf{ - F(y) = constant}}\).

(a)

Let,

\(\begin{array}{c}{\bf{v = y'}}\\{\bf{v' = y''}}{\bf{.y'' = f(y)}}\end{array}\)

Then the system is:

\(\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = v}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = f(y)}}\end{array}\)

Yes,\(y'' = f\left( y \right)\) as an equivalent first-order system.

(b)

Here,

\(\begin{array}{c}\frac{{{\bf{dv}}}}{{{\bf{dy}}}}{\bf{ = }}\frac{{{\bf{f(y)}}}}{{\bf{v}}}\\\int {{\bf{vdv}}} {\bf{ = }}\int {{\bf{f(y)dy}}} \\\frac{{{{\bf{v}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ = F(y) + K}}\end{array}\)

Hence, \(\frac{{{{\bf{v}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ = F(y) + K}}\)the solutions to the vy-phase plane equation for the system are in part.

Now put \({\bf{v = y'}}\)then;

\(\frac{{{\bf{y'(t}}{{\bf{)}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - F(y) = constant}}\)

Therefore, the verifying solution is \(\frac{{{\bf{y'(t}}{{\bf{)}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - F(y) = constant}}\)