【General guidance】The answer provided below has been developed in a clear step by step manner.

Step1/3

(a) To find the extremal function of the given integral, we need to use the Euler-Lagrange equation \( \mathrm{\frac{{d}}{{\left.{d}{x}\right.}}{\left[\frac{{∂{f}}}{{∂{y}'}}\right]}-\frac{{∂{f}}}{{∂{y}}}={0}} \).

,where \( \mathrm{{f}={\left(\frac{{1}}{{2}}\right)}{x}^{{2}}{\left({y}'\right)}^{{2}}+{y}^{{2}}+{x}^{{2}}{y}} \).

Let's start by computing the partial derivative of \( \mathrm{{f}} \) with respect to \( \mathrm{{y}'} \):

\( \mathrm{\frac{{∂{f}}}{{∂{y}'}}={x}^{{2}}{y}'} \)

Now let's compute the derivative of this with respect to \( \mathrm{{x}} \):

\( \mathrm{\frac{{d}}{{\left.{d}{x}\right.}}{\left({x}^{{2}}{y}'\right)}={2}{x}{y}'+{x}^{{2}}{y}{''}} \)

Next, we need to compute the partial derivative of \( \mathrm{{f}} \) with respect to \( \mathrm{{y}} \):

\( \mathrm{\frac{{∂{f}}}{{∂{y}}}={2}{y}+{x}^{{2}}} \).

Now we can plug these into the Euler-Lagrange equation:

\( \mathrm{{2}{x}{y}'+{x}^{{2}}{y}{''}-{\left({2}{y}+{x}^{{2}}\right)}={0}} \)

\( \mathrm{\Rightarrow{x}^{{2}}{y}{''}+{2}{x}{y}'-{2}{y}={x}^{{2}}} \).

This is a second-order linear differential equation.

Explanation: Please refer to solution in this step.

Step2/3

To solve this equation we substitute, \( \mathrm{{x}={e}^{{z}}} \). Then we get,

\( \mathrm{{\left[{D}{\left({D}-{1}\right)}+{2}{D}-{2}\right]}{y}={e}^{{{2}{z}}}} \) , where \( \mathrm{{D}\equiv\frac{{d}}{{{\left.{d}{z}\right.}}}} \).

Then, \( \mathrm{{\left({D}^{{2}}+{D}-{2}\right)}{y}={e}^{{{2}{z}}}} \).

Auxiliary equation of the corresponding homogeneous equation is \( \mathrm{{r}^{{2}}+{r}-{2}={0}} \), where \( \mathrm{{y}={e}^{{{r}{x}}}} \) is the trial solution of the corresponding homogeneous differential equation.

Then, \( \mathrm{{\left({r}+{2}\right)}{\left({r}-{1}\right)}={0}} \), which implies \( \mathrm{{r}={1},-{2}} \).

Then the complementary function is \( \mathrm{{y}_{{c}}={A}{e}^{{z}}+{B}{e}^{{-{2}{z}}}} \).

Now, \( \mathrm{{y}_{{p}}=\frac{{1}}{{{D}^{{2}}+{D}-{2}}}{e}^{{{2}{z}}}=\frac{{{e}^{{{2}{z}}}}}{{{2}^{{2}}+{2}-{2}}}=\frac{{1}}{{4}}{e}^{{{2}{z}}}} \).

Therefore, the solution of the differential equation is given by

\( \mathrm{{y}={y}_{{c}}+{y}_{{p}}={A}{e}^{{z}}+{B}{e}^{{-{2}{z}}}+\frac{{1}}{{4}}{e}^{{{2}{z}}}} \).

Finally substituting \( \mathrm{{e}^{{z}}={x}} \) we get,

\( \mathrm{{y}={A}{x}+{B}{x}^{{-{2}}}+\frac{{1}}{{4}}{x}^{{2}}} \).

This is the required extremal of the given function, where \( \mathrm{{A}} \) and \( \mathrm{{B}} \) are arbitrary constants.

Explanation: Please refer to solution in this step.

Step3/3

Explanation

To find the extremal function of the given integral, we need to use the Euler-Lagrange equation \( \mathrm{\frac{{d}}{{\left.{d}{x}\right.}}{\left[\frac{{∂{f}}}{{∂{y}'}}\right]}-\frac{{∂{f}}}{{∂{y}}}={0}} \).

(b) where \( \mathrm{{f}={\left({y}'\right)}^{{2}}-{y}^{{2}}+{2}{y}{\sec{{\left({x}\right)}}}} \)

Let's start by computing the partial derivative of \( \mathrm{{f}} \) with respect to \( \mathrm{{y}'} \):

\( \mathrm{\frac{{∂{f}}}{{∂{y}'}}={2}{y}'} \)

Now let's compute the derivative of this with respect to \( \mathrm{{x}} \):

\( \mathrm{\frac{{d}}{{\left.{d}{x}\right.}}{\left({2}{y}'\right)}={2}{y}{''}} \)

Next, we need to compute the partial derivative of \( \mathrm{{f}} \) with respect to \( \mathrm{{y}} \):

\( \mathrm{\frac{{∂{f}}}{{∂{y}}}=-{2}{y}} \)

Finally, we need to compute the partial derivative of \( \mathrm{{f}} \) with respect to \( \mathrm{{x}} \):

\( \mathrm{\frac{{∂{f}}}{{∂{x}}}={2}{y}{\sec{{\left({x}\right)}}}{\tan{{\left({x}\right)}}}} \)

Now we can plug these into the Euler-Lagrange equation:

\( \mathrm{{2}{y}{''}+{2}{y}-{2}{y}{\sec{{\left({x}\right)}}}{\tan{{\left({x}\right)}}}={0}} \).

This equation cannot be solve by analytical method.

Explanation: Please refer to solution in this step.

Find the extremal function of the following: a) Integral from a to b of [1/2 x^2 (y')^2 + y^2 + x^2y] dx b) Integral from a to b [(y')^2 - y^2 + 2y sec x] dx

Public Answer

Find the extremal function of the following: a) Integral from a to b of [1/2 x^2 (y')^2 + y^2 + x^2y] dx

b) Integral from a to b [(y')^2 - y^2 + 2y sec x] dx