Question

Sketch the solution to the initial value problem \(\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = 2y - 2yt,}}\,\,\,\,\,\,\,\,\,\,{\bf{y}}\left( {\bf{0}} \right){\bf{ = 3}}\)and determine its maximum value.

Short answer

The maximum value is at t=1.

The graph is drawn below.

Step-by-step solution

Step 1:Cross multiplication for simplification

Given\[\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = 2y - 2yt}}......\left( {\bf{1}} \right)\]

Cross multiplication on both sides in the equation \(\left( 1 \right)\)

\[\begin{array}{l}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = 2y}}\left( {{\bf{1 - t}}} \right)\\\frac{{{\bf{dy}}}}{{\bf{y}}}{\bf{ = 2}}\left( {{\bf{1 - t}}} \right){\bf{dt}}\end{array}\]

Step 2:Apply integration

Taking integration of both sides in the above equation

\[\int {\frac{{{\bf{dy}}}}{{\bf{y}}}} {\bf{ = 2}}\int {\left( {{\bf{1 - t}}} \right){\bf{dt}}} \]

Simplify the above equation,

\[{\bf{ln}}\left( {\bf{y}} \right){\bf{ = 2t - }}{{\bf{t}}^{\bf{2}}}{\bf{ + c}}......\left( {\bf{2}} \right)\]

Here c is the integration constant.

Step 3:Put the initial condition \({\bf{y}}\left( {\bf{0}} \right){\bf{ = 3}}\)

Substituting\(y\left( 0 \right) = 3\)in the equation\(\left( 2 \right)\)

\[\begin{array}{l}{\bf{ln}}\left( {\bf{3}} \right){\bf{ = 2}}\left( {\bf{0}} \right){\bf{ - 0 + c}}\\{\bf{ln}}\left( {\bf{3}} \right){\bf{ = c}}\end{array}\]

Step 4:Simplify for the value of c

Put the value of c in the equation\(\left( 2 \right)\)

\[{\bf{ln}}\left( {\bf{y}} \right){\bf{ = 2t - }}{{\bf{t}}^{\bf{2}}}{\bf{ + ln}}\left( {\bf{3}} \right)\]

Simplify the above equation.

\[\begin{array}{c}{\bf{ln}}\left( {\bf{y}} \right){\bf{ - ln}}\left( {\bf{3}} \right){\bf{ = 2t - }}{{\bf{t}}^{\bf{2}}}\\{\bf{ln}}\left( {\frac{{\bf{y}}}{{\bf{3}}}} \right){\bf{ = 2t - }}{{\bf{t}}^{\bf{2}}}\\\frac{{\bf{y}}}{{\bf{3}}}{\bf{ = }}{{\bf{e}}^{{\bf{2t - }}{{\bf{t}}^{\bf{2}}}}}\\{\bf{y = 3}}{{\bf{e}}^{{\bf{2t - }}{{\bf{t}}^{\bf{2}}}}}\end{array}\]

Find the maximum value

\[\begin{array}{l}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = 0}}\\\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = 3}}\left( {{\bf{2 - 2t}}} \right){{\bf{e}}^{{\bf{2t - }}{{\bf{t}}^{\bf{2}}}}}{\bf{ = 0}}\end{array}\]

Using the zero-factor principal: if\(ab = 0\)then\(a = 0\)or\(b = 0\),

\[\begin{array}{c}{\bf{3}}\left( {{\bf{2 - 2t}}} \right){\bf{ = 0}}\\{\bf{t = 0}}\end{array}\]

Or

\[\begin{array}{c}{{\bf{e}}^{{\bf{2t - }}{{\bf{t}}^{\bf{2}}}}}{\bf{ = 0}}\\{\bf{2t - }}{{\bf{t}}^{\bf{2}}}{\bf{ = 1}}\\{{\bf{t}}^{\bf{2}}}{\bf{ - 2t + 1 = 0}}\\{\left( {{\bf{t - 1}}} \right)^{\bf{2}}}{\bf{ = 0}}\\{\bf{t = \pm 1}}\end{array}\]

\[\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = 3}}\left( {{\bf{2 - 2t}}} \right){{\bf{e}}^{{\bf{2t - }}{{\bf{t}}^{\bf{2}}}}}\\\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = 3}}\left( {{\bf{ - 2}}} \right){{\bf{e}}^{{\bf{2t - }}{{\bf{t}}^{\bf{2}}}}}{\bf{ + 3}}{\left( {{\bf{2 - 2t}}} \right)^{\bf{2}}}{{\bf{e}}^{{\bf{2t - }}{{\bf{t}}^{\bf{2}}}}}\end{array}\]

At\(t = 1\)

\[\begin{array}{l}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - 6e + 3}}\left( {{\bf{2 - 2}}} \right){\bf{e}}\\\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - 6e < 0}}\end{array}\]

It has a maximum at\(t = 1\)

Therefore the maximum value at t=1.