Engineering Mathematics Test 7

Concept:

The Laplace transform of a general exponential signal is given by:

\(L[e^{-at}]\longleftrightarrow \frac{1}{s+a}\)

where 'a' is any positive integer.

Calculation:

Given, The Laplace transform is given as:

F(s) = \(\frac{2}{s+1}\)

The inverse Laplace transform will be:

\(\frac{2}{s+1}\longleftrightarrow2e^{-t}\)

One dimensional wave equation: \(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

Two-dimensional wave equation: \(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}\)

Important:

Heat equation: \(\frac{{\partial u}}{{\partial t}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

Laplace equation: \(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} = 0\)

Concept:

A second-order linear partial differential equation in two variables with constant coefficients in the form of

\({\rm{A}}\frac{{{\partial ^2}{\rm{u}}}}{{\partial {{\rm{x}}^2}}} + {\rm{B}}\frac{{{\partial ^2}{\rm{u}}}}{{\partial {\rm{x}}\partial {\rm{y}}}} + {\rm{C}}\frac{{{\partial ^2}{\rm{u}}}}{{\partial {{\rm{y}}^2}}} + {\rm{D}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{x}}}} + {\rm{E}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{y}}}} + {\rm{Fu}} = {\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right)\) can be called,

1. Elliptical if discriminant \({{\rm{B}}^2} - 4{\rm{AC}} < 0\)
2. Parabolic if discriminant \({{\rm{B}}^2} - 4{\rm{AC}} = 0\)
3. Hyperbolic if discriminant \({{\rm{B}}^2} - 4{\rm{AC}} > 0\)

Where A, B, C, D, E, F are constants. The nature of the equation is independent of the values of the constants D, E and F.

Calculation:

The given partial differential equation is to be elliptical in nature. So, the discriminant should be less the zero.

\(\therefore {\rm{\;The\;discriminant\;}} = {\left( { - {\rm{a}}} \right)^2} - 4 \times 7 \times 7 < 0\;\;\;\therefore {{\rm{a}}^2} < 196{\rm{\;}}\therefore {\rm{\;}} - 14 < a < 14\)

Hence, the partial differential equation will be elliptical in nature for any real value of \( - 14 < a < 14\) and for any the values of c because the nature of the equation is independent of the value c.

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)

where

\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;\;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;\;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)

An even function is any function f such that f(-x) = f(x)

Example: cos x, sec x, x², x⁴, x⁶ …….., x^-2, x^-4 ……..

An odd function is any function f such that f(-x) = -f(x)

Example: sin x, tan x, cosec x, cot x, n, x³ ……., x^-1, x^-3 ……..

\(\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^L f\left( x \right)dx,\;\;when\;f\left( x \right)\;is\;an\;even\;function}\\ {0,\;\;when\;f\left( x \right)\;is\;an\;odd\;function} \end{array}} \right.\)

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

\({a_o} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)dx\)

\({a_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx\)

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

\({b_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx\)

Concept:

Laplace Transformation:

Laplace Transformation formula of \({\rm{L}}\left[ 1 \right] = \frac{1}{{\rm{s}}}{\rm{\;and\;L}}\left[ {\cos {\rm{at}}} \right] = \frac{{\rm{s}}}{{{{\rm{s}}^2} + {{\rm{a}}^2}{\rm{\;}}}}{\rm{\;where\;s}} > 0\).

Calculation:

\({\rm{L}}\left[ {{\rm{f}}\left( {\rm{t}} \right)} \right] = {\rm{L}}\left[ {{{\cos }^2}4{\rm{t\;}}} \right] = {\rm{L}}\left[ {\frac{1}{2}\left( {1 + \cos 8{\rm{t}}} \right)} \right] = \frac{1}{2} \times \left[ {{\rm{L}}\left[ 1 \right] + {\rm{L}}\left[ {\cos 8{\rm{t}}} \right]} \right] = \frac{1}{2}\left[ {\frac{1}{{\rm{s}}} + \frac{{\rm{s}}}{{{{\rm{s}}^2} + {8^2}{\rm{\;}}}}} \right]\)

Hence, the Laplace Tranformation of f(t)is \(\frac{1}{2}\left[ {\frac{1}{{\rm{s}}} + \frac{{\rm{s}}}{{{{\rm{s}}^2} + 64{\rm{\;}}}}} \right]\)

Confusion Point:

\({\rm{L}}\left[ {\sin {\rm{at}}} \right] = \frac{{\rm{a}}}{{{{\rm{s}}^2} + {{\rm{a}}^2}{\rm{\;}}}}{\rm{\;where\;s}} > 0\)

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} 0&{ - \pi < x < 0}\\ 1&{0 < x < \pi } \end{array}} \right.\)

\({a_0} = \frac{1}{\pi }\mathop \smallint \limits_{ - \pi }^\pi f\left( x \right)dx\)

\(= \frac{1}{\pi }\left[ {\mathop \smallint \limits_{ - \pi }^0 0\;dx + \mathop \smallint \limits_0^\pi 1\;dx} \right]\)

\(= \frac{1}{\pi }\left[ {\mathop \smallint \limits_0^\pi 1\;dx} \right] = \frac{1}{\pi }\left( \pi \right) = 1\)

\({a_n} = \frac{1}{\pi }\mathop \smallint \limits_{ - \pi }^\pi f\left( x \right)\cos nx\;dx\)

\(= \frac{1}{\pi }\mathop \smallint \limits_{ - \pi }^\pi \cos nx\;dx\)

\(= \frac{1}{\pi }{\left[ {\frac{{\sin nx}}{n}} \right]_0}^\pi = 0\)

\({b_n} = \frac{1}{\pi }\mathop \smallint \limits_{ - \pi }^\pi f\left( x \right)\sin nx\;dx\)

\(= \frac{1}{\pi }\mathop \smallint \limits_0^\pi \sin nx\;dx\)

\(= \frac{1}{\pi }{\left[ {\frac{{ - \cos nx}}{n}} \right]_0}^\pi \)

\(= \frac{1}{\pi }\left[ {\frac{{ - \cos n\pi }}{n} + \frac{{\cos 0}}{n}} \right]\)

\(= \frac{{1 - {{\left( { - 1} \right)}^n}}}{{n\pi }}\)

\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^ \propto [{a_n}\cos nx + {b_n}\sin nx]\)

\(= \frac{1}{2} + \mathop \sum \limits_{n = 1}^ \propto \frac{{1 - {{\left( { - 1} \right)}^n}}}{{n\pi }}\sin nx\)

\(= \frac{1}{2} + \frac{2}{\pi }\sin x + \frac{2}{{3\pi }}\sin 3\;x + \frac{2}{{5\pi }}\sin 5x + \; - - - \)

Co-efficient of first harmonic \(= \frac{2}{\pi }\)

Co-efficient of third harmonic \(= \frac{2}{{3\pi }}\)

\(Ratio = \frac{{\frac{2}{\pi }}}{{\frac{2}{{3\pi }}}} = 3\)

\(\begin{array}{l}L\left\{ {\cos t} \right\} = \frac{s}{{{s^2} + 1}}\\L\left\{ {{e^{ - 2t}}\cos t} \right\} = \frac{{s + 2}}{{{{\left( {s + 2} \right)}^2} + 1}} = \frac{{s + 2}}{{{s^2} + 4s + 5}}\end{array}\)

\(L\left\{ {t{e^{ - 2t}}\cos t} \right\} = \frac{{ - d}}{{ds}}\left( {\frac{{s + 2}}{{{s^2} + 4s + 5}}} \right)\)

\(= - \left[ {\frac{{\left( {{s^2} + 4s + 5} \right)\left( 1 \right) - \left( {s + 2} \right)\left( {2s + 4} \right)}}{{{{\left( {{s^2} + 4s + 5} \right)}^2}}}} \right]\)

\(= \frac{{2{s^2} + 8 + 8s - {s^2} - 4s - 5}}{{{{\left( {{s^2} + 4s + 5} \right)}^2}}}\)

\(= \frac{{\left( {{s^2} + 4s + 3} \right)}}{{{{\left( {{s^2} + 4s + 5} \right)}^2}}}\)

\(F\left( s \right) = L\left\{ {\mathop \smallint \limits_0^t t{e^{ - 2t}}\cos t} \right\} = \frac{{\left( {{s^2} + 4s + 3} \right)}}{{s{{\left( {{s^2} + 4s + 5} \right)}^2}}}\)

By final value theorem,

\(\mathop {{\rm{lt}}}\limits_{s \to 0} s.F\left( s \right) = \mathop {{\rm{lt}}}\limits_{s \to 0} \frac{{s\left( {{s^2} + 4s + 3} \right)}}{{s{{\left( {{s^2} + 4s + 5} \right)}^2}}} = \frac{3}{{25}}\)

Concept:

Solution of Heat Equation:

The general form of the one-dimensional heat equation is \(\frac{{\partial {\rm{u}}}}{{\partial {\rm{t}}}} = {{\rm{\beta }}^2}\frac{{{\partial ^2}{\rm{u}}}}{{\partial {{\rm{x}}^2}}}\) and this equation can be solved by applying method separation. The closed form of the solution is given by,

\({\rm{u}}\left( {{\rm{x}},{\rm{t}}} \right) = {\rm{\;}}\left( {{\rm{Acospx}} + {\rm{Bsinpx}}} \right){{\rm{e}}^{ - {{\rm{\beta }}^2}{{\rm{p}}^2}{\rm{t}}}}\) where A and B are constants and \( - {{\rm{p}}^2}\) is the separation constant.

By using boundary conditions and initial conditions values of A, B and p can be calculated.

Calculation:

Given PDE can be rearranged as, \(\frac{{\partial {\rm{h}}}}{{\partial {\rm{t}}}} = {\left( {\frac{1}{{1.6}}} \right)^2}\frac{{{\partial ^2}{\rm{h}}}}{{\partial {{\rm{x}}^2}}}\)

Hence, the solution can be written as, \({\rm{h}}\left( {{\rm{x}},{\rm{t}}} \right) = {\rm{\;}}\left( {{\rm{Acospx}} + {\rm{Bsinpx}}} \right){{\rm{e}}^{ - {{\left( {\frac{1}{{1.6}}} \right)}^2}{{\rm{p}}^2}{\rm{t}}}}\)

Now, the boundary condition is given as \({\rm{h}}\left( {0,{\rm{t}}} \right) = {\rm{\;}}0{\rm{\;\;}}\therefore {\rm{A}} = 0\therefore {\rm{h}}\left( {{\rm{x}},{\rm{t}}} \right) = {\rm{\;}}\left( {{\rm{Bsinpx}}} \right){{\rm{e}}^{ - {{\left( {\frac{1}{{1.6}}} \right)}^2}{{\rm{p}}^2}{\rm{t}}}}\) 

And, \({\rm{h}}\left( {2,{\rm{t}}} \right) = {\rm{\;}}0{\rm{\;\;}}\therefore {\rm{sin}}2{\rm{p}} = 0{\rm{\;}} \Rightarrow {\rm{p}} = \frac{{{\rm{n\pi }}}}{2}{\rm{\;\;\;\;\;}}\therefore {\rm{h}}\left( {{\rm{x}},{\rm{t}}} \right) = {\rm{\;}}\left( {{\rm{Bsin}}\frac{{{\rm{n\pi }}}}{2}{\rm{x}}} \right){{\rm{e}}^{ - {{\left( {\frac{1}{{1.6}}} \right)}^2}{{\left( {\frac{{{\rm{n\pi }}}}{2}} \right)}^2}{\rm{t}}}}\) 

Now ignoring the general form of the solution, and applying the initial condition given as \({\rm{h}}\left( {{\rm{x}},0} \right) = {\rm{sin}}3.2{\rm{x}}\)

\(\therefore {\rm{Bsin}}\frac{{{\rm{n\pi }}}}{2}{\rm{x}} = {\rm{sin}}3.2{\rm{x\;\;\;}}\therefore {\rm{Comapring\;both\;sides}},{\rm{\;B}} = 1{\rm{\;and\;\;}}\frac{{{\rm{n\pi }}}}{2} = 3.2\)

So, the final form of the solution is, \({\rm{h}}\left( {{\rm{x}},{\rm{t}}} \right) = {\rm{\;}}\left( {{\rm{sin}}3.2{\rm{x}}} \right){{\rm{e}}^{ - {{\left( {\frac{1}{{1.6}}} \right)}^2}{{\left( {3.2} \right)}^2}{\rm{t}}}} = {{\rm{e}}^{ - 4{\rm{t}}}}{\rm{sin}}3.2{\rm{x}}\)

Tips and Tricks:

The solution of 2D wave equations, heat equations and Laplace equations are lengthy. If possible, it’s always recommended to go by verifying options. Consider the fourth option.

\({\rm{h}}\left( {{\rm{x}},{\rm{t}}} \right) = {{\rm{e}}^{ - 4{\rm{t}}}}{\rm{sin}}3.2{\rm{x\;}}\therefore {\rm{\;}}\frac{{\partial {\rm{h}}}}{{\partial {\rm{t}}}} = - 4{{\rm{e}}^{ - 4{\rm{t}}}}{\rm{sin}}3.2{\rm{x\;\;and\;}}\frac{{\partial {\rm{h}}}}{{\partial {\rm{x}}}} = 3.2{{\rm{e}}^{ - 4{\rm{t}}}}{\rm{cos}}3.2{\rm{x\;}} \Rightarrow {\rm{\;}}\frac{{{\partial ^2}{\rm{h}}}}{{\partial {{\rm{x}}^2}}} = {3.2^2}{{\rm{e}}^{ - 4{\rm{t}}}}{\rm{sin}}3.2{\rm{x\;\;}}\)

\(\Rightarrow {\rm{\;}}\frac{{{\partial ^2}{\rm{h}}}}{{\partial {{\rm{x}}^2}}} = 10.24 \times {{\rm{e}}^{ - 4{\rm{t}}}}{\rm{sin}}3.2{\rm{x\;}} = 2.56 \times 4{{\rm{e}}^{ - 4{\rm{t}}}}{\rm{sin}}3.2{\rm{x}} = 2.56\frac{{\partial {\rm{h}}}}{{\partial {\rm{t}}}}{\rm{\;\;}}\therefore \frac{{{\partial ^2}{\rm{h}}}}{{\partial {{\rm{x}}^2}}} = 2.56\frac{{\partial {\rm{h}}}}{{\partial {\rm{t}}}}\left( {{\rm{checked\;as\;given}}} \right)\)

Concept:

Solution of Linear Homogeneous Equation:

If an auxiliary equation of a linear homogeneous equation has n roots out of which one real root occurs k times and the remaining (n – k) numbers of the roots are distinct roots viz. \({{\rm{m}}_{{\rm{k}} + 1}},{{\rm{m}}_{{\rm{k}} + 2}}, \ldots ,{{\rm{m}}_{\rm{n}}}\) then the solution of the differential equation is given by,

\(\left( {{{\rm{c}}_1} + {{\rm{c}}_2}{\rm{x}} + {{\rm{c}}_3}{{\rm{x}}^2} + \ldots + {{\rm{c}}_{\rm{k}}}{{\rm{x}}^{{\rm{k}} - 1}}} \right){{\rm{e}}^{{{\rm{m}}_{\rm{k}}}{\rm{x}}}} + {{\rm{c}}_{{\rm{k}} + 1}}{{\rm{e}}^{{{\rm{m}}_{{\rm{k}} + 1}}{\rm{x}}}} + {{\rm{c}}_{{\rm{k}} + 2}}{{\rm{e}}^{{{\rm{m}}_{{\rm{k}} + 2}}{\rm{x}}}} + \ldots + {{\rm{c}}_{\rm{n}}}{{\rm{e}}^{{{\rm{m}}_{\rm{n}}}{\rm{x}}}}\)

Calculation:

The given differential equation can be written as, \(\left( {{{\rm{D}}^3} - 3{{\rm{D}}^2} + 4} \right){\rm{y}} = 0\)

So, the auxiliary equation is, \({{\rm{m}}^3} - 3{{\rm{m}}^2} + 4 = 0{\rm{\;}} \Rightarrow \left( {{\rm{m}} + 1} \right){\left( {{\rm{m}} - 2} \right)^2} = 0 \Rightarrow {\rm{m}} = - 1,2,2\)

Hence the solution is, \({\rm{y}} = \left( {{{\rm{c}}_1} + {{\rm{c}}_2}{\rm{x}}} \right){{\rm{e}}^{2{\rm{x}}}} + {{\rm{c}}_3}{{\rm{e}}^{ - {\rm{x}}}}{\rm{\;\;}} \Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = 2{{\rm{e}}^{2{\rm{x}}}}\left( {{{\rm{c}}_1} + {{\rm{c}}_2}{\rm{x}}} \right) + {\rm{\;}}{{\rm{e}}^{2{\rm{x}}}}{{\rm{c}}_2} - {{\rm{c}}_3}{{\rm{e}}^{ - {\rm{x}}}}{\rm{\;\;}}\)

Given, \({\rm{y}} = 0{\rm{\;at\;x}} = 0{\rm{\;\;}}\therefore {{\rm{c}}_1} = - {{\rm{c}}_3}{\rm{\;\;and\;\;y}} = 1{\rm{\;at\;x}} = 1{\rm{\;\;\;}}\therefore 1 = \left( {{{\rm{c}}_1} + {{\rm{c}}_2}} \right){{\rm{e}}^2} + \frac{{{{\rm{c}}_3}}}{{\rm{e}}}{\rm{\;\;}}\)

And, \(\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = 0{\rm{\;at\;x}} = 0{\rm{\;\;\;}}\therefore 0 = 2\left( {{{\rm{c}}_1}} \right) + {{\rm{c}}_2} - {{\rm{c}}_3} \Rightarrow {{\rm{c}}_2} = {{\rm{c}}_3} - 2{{\rm{c}}_1} = - {{\rm{c}}_1} - 2{{\rm{c}}_1} = - 3{{\rm{c}}_1}\)

Now, \(1 = \left( {{{\rm{c}}_1} + {{\rm{c}}_2}} \right){{\rm{e}}^2} + \frac{{{{\rm{c}}_3}}}{{\rm{e}}} \Rightarrow 1 = \left( {{{\rm{c}}_1} - 3{{\rm{c}}_1}} \right){{\rm{e}}^2} - \frac{{{{\rm{c}}_1}}}{{\rm{e}}} \Rightarrow {\rm{e}} = - 2{{\rm{c}}_1}{{\rm{e}}^3} - {{\rm{c}}_1} \Rightarrow {{\rm{c}}_1} = \frac{{ - {\rm{e}}}}{{1 + 2{{\rm{e}}^3}}}\) 

\(\therefore {{\rm{c}}_3} = \frac{{\rm{e}}}{{1 + 2{{\rm{e}}^3}}}{\rm{\;and\;}}{{\rm{c}}_2} = \frac{{3{\rm{e}}}}{{1 + 2{{\rm{e}}^3}}}\)

Hence the solution is, \({\rm{y}} = \frac{{\rm{e}}}{{1 + 2{{\rm{e}}^3}}}\left( { - 1 + 3{\rm{x}}} \right){{\rm{e}}^{2{\rm{x}}}} + \frac{{\rm{e}}}{{1 + 2{{\rm{e}}^3}}}{{\rm{e}}^{ - {\rm{x}}}}\)

\(\therefore {\rm{y}}\left( {\frac{1}{3}} \right) = \frac{{\rm{e}}}{{1 + 2{{\rm{e}}^3}}}{{\rm{e}}^{ - {\rm{\;}}\frac{1}{3}}} = 0.047\)

Concept:

The approach for obtaining the solutions of the differential equations in the form of \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{f}}\left( {\rm{y}} \right)\) is to multiply both sides by \(2\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) and integrate both sides two times. Then using the boundary conditions find out the unknown constants associated with the solution.

Calculation:

Given the differential equation is \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} = 7{{\rm{y}}^{\frac{2}{5}}}\)

Multiplying both sides by \(2\frac{{{\rm{dy}}}}{{{\rm{dt}}}}\) we get, \(2\frac{{{\rm{dy}}}}{{{\rm{dt}}}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} = 14{{\rm{y}}^{\frac{2}{5}}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}} \Rightarrow \frac{{\rm{d}}}{{{\rm{dt}}}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dt}}}}} \right)^2} = 14{{\rm{y}}^{\frac{2}{5}}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}}\)

Integrating both sides with respect to t we get, \({\left( {\frac{{{\rm{dy}}}}{{{\rm{dt}}}}} \right)^2} = \frac{{14{{\rm{y}}^{\frac{7}{5}}}}}{{7/5}} + {\rm{c}} = 10{{\rm{y}}^{\frac{7}{5}}} + {\rm{c}}\)

Given, \({\rm{y\;}} = {\rm{\;}}1{\rm{\;and\;}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}} = \sqrt {10} {\rm{\;\;at\;t}} = 0{\rm{\;\;\;}}\therefore 10 = 10 \times {\left( 1 \right)^{\frac{7}{5}}} + {\rm{c}} \Rightarrow {\rm{c}} = 0{\rm{\;}}\)  

\(\therefore {\left( {\frac{{{\rm{dy}}}}{{{\rm{dt}}}}} \right)^2} = 10{{\rm{y}}^{\frac{7}{5}}} \Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dt}}}} = \sqrt {10} {\rm{\;}}{{\rm{y}}^{0.7}} \Rightarrow \smallint {{\rm{y}}^{ - 0.7}}{\rm{dy}} = \smallint \sqrt {10{\rm{\;}}} {\rm{dt}} + {{\rm{c}}_1}{\rm{\;}} \Rightarrow \frac{{{{\rm{y}}^{0.3}}}}{{0.3}} = \sqrt {10} {\rm{\;t}} + {{\rm{c}}_1}\)

Given, \({\rm{y\;}} = {\rm{\;}}1{\rm{\;at\;t}} = 0{\rm{\;\;\;\;}}\therefore {{\rm{c}}_1} = \frac{{10}}{3}{\rm{\;\;\;\;}}\therefore {\rm{\;}}\frac{{{{\rm{y}}^{0.3}}}}{{0.3}} = \sqrt {10} {\rm{\;t}} + \frac{{10}}{3}{\rm{\;}} \Rightarrow {{\rm{y}}^{0.3}} = 1 + 0.3\sqrt {10} {\rm{\;t\;\;}}\)

\(\therefore {\rm{y}} = {\left( {1 + 0.3\sqrt {10} {\rm{\;t}}} \right)^{10/3}}{\rm{\;\;\;\;Hence}},{\rm{\;y}}\left( {\sqrt {10} } \right) = {\rm{\;}}{\left( {1 + 0.3\sqrt {10} {\rm{\;}} \times \sqrt {10} } \right)^{10/3}} = 101.59\)