Question

Find the analytic solution of the initial-value problem in Example 1 . Compare the actual values of \({\bf{y}}\left( {{\bf{0}}.{\bf{2}}} \right),{\rm{ }}{\bf{y}}\left( {{\bf{0}}.{\bf{4}}} \right),{\rm{ }}{\bf{y}}\left( {{\bf{0}}.{\bf{6}}} \right),\)and \({\bf{y}}\left( {{\bf{0}}.8} \right)\)with the approximations \({\bf{y}}\left( {{\bf{0}}.{\bf{2}}} \right),{\rm{ }}{\bf{y}}\left( {{\bf{0}}.{\bf{4}}} \right),{\rm{ }}{\bf{y}}\left( {{\bf{0}}.{\bf{6}}} \right),\)and \({{\bf{y}}_4}.\)

Short answer

The approximation value of \(y\left( {0.2} \right){\rm{, y}}\left( {0.4} \right){\rm{, y}}\left( {0.6} \right)\)and \(y\left( {0.8} \right)\) are \(1.02140000\),\(1.09181796\),\(1.22210646\)and \(1.42552788\)

Step-by-step solution

Definition of analytic solution

• An analytical answer involves rephrasing the problem in a comprehensible manner and calculating the exact solution.
• In general, the analytical method is preferred since it is quicker and provides more precise results.
• Nonetheless, due to time or hardware constraints, we may be forced to use a numerical method.

To find the analytic solution

We have the first order differential equation

\({y^\prime } = x + y - 1\)

Which is

\({y^\prime } - y = x - 1\)

With the initial condition

\(y(x = 0) = 1\)

Which is in the differential equation form

\(\frac{{dy}}{{dx}} + P(x)y = Q(x)\)

and we have to solve it as the following technique :

First, we have to find an integrating factor as

\({\rm{ I}}{\rm{.F }} = {e^{\int P (x)dx}}\)

\(\begin{aligned}{l} = {e^{ - \int d x}}\\ = {e^{ - x}}\end{aligned}\)

Second, multiply both sides of the given differential equation by this integrating factor, then we have

\({e^{ - x}}{y^\prime } - {e^{ - x}}y = (x - 1){e^{ - x}}\)

Now, if we backed one step, we would have

\(\begin{aligned}{l}\frac{d}{{dx}}\left( {{e^{ - x}}y} \right) = (x - 1){e^{ - x}}\\\int d \left( {{e^{ - x}}y} \right) = \int {(x - 1)} {e^{ - x}}dx{e^{ - x}}y = \int {(x - 1)} {e^{ - x}}dx\end{aligned}\)

Integrating \(\int {(x - 1)} {e^{ - x}}dx\) to find the solutions.

This integration \(\int {(x - 1)} {e^{ - x}}dx\)can be made using integration by parts as the following

If we set \(u = \left( {x - 1} \right),\)then we have \(d{\rm{ }}u = d{\rm{ }}x,\)and if we set \(dv = {e^{ - x}}dx,\)then we have \(v = - {e^{ - x}},\) then we can solve integration as

\(\int {(x - 1)} {e^{ - x}}dx = u \times v - \int v du\)

\(\begin{aligned}{l} = (1 - x){e^{ - x}} + \int {{e^{ - x}}} dx\\ = (1 - x){e^{ - x}} - {e^{ - x}} + c\\ = - x{e^{ - x}} + c\end{aligned}\)

Then we have

\({e^{ - x}}y = - x{e^{ - x}} + c\) (1)

After that, to find the value of constant \(c,\)we have to apply the condition \((x,y) = (0,1)\)into equation \(\left( 1 \right),\)then we have

\(\begin{aligned}{l}{e^0} \times 1 = 0 + c\\1 = 0 + c\\c = 1\end{aligned}\)

Finally, substitute with the value of constant \(c\)into equation (1), then we have

\({e^{ - x}}y = - x{e^{ - x}} + 1\)

Then we have

\(y = - x + {e^x}\) (2)

Is the solution of equation.

Determining the actual values from its approximated values.

Now using equation(2), we can obtain the required actual values as the following

The actual value \(y(0.2)\) is

\(y(0.2) = - 0.2 + {e^{0.2}}\)

=1.02140275

The approximated value from example (1) is

\({y_1} = 1.02140000\)

The actual value \(y(0.4)\)is

\(y(0.4) = - 0.4 + {e^{0.4}}\)

\( = 1.09182470\)

The approximated value from example (1) is

\({y_2} = 1.09181796\)

The actual value \(y(0.6)\)is

\(y(0.4) = - 0.6 + {e^{0.6}}\)

\( = 1.22211880\)

The approximated value from example (1) is

\({y_3} = 1.22210646\)

The actual value \(y(0.8)\)is

\(y(0.4) = - 0.8 + {e^{0.8}}\)

\( = 1.42554093\)

The approximated value from example (1) is

\({y_4} = 1.42552788\)

The analytic solution is \(y = - x + {e^x}\)

Therefore the actual value of \(y\left( {0.2} \right){\rm{, y}}\left( {0.4} \right){\rm{, y}}\left( {0.6} \right)\)and \(y\left( {0.8} \right)\) are \(1.02140000\),\(1.09181796\),\(1.22210646\)and \(1.42552788\)respectively.