Engineering Mathematics Test 6

The order of differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative accruing in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

\({\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^{\frac{1}{2}}} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^{\frac{1}{3}}}\)

Highest derivative in the given differential equation is 4

Hence order is 4.

To find the degrees, we need to change the differential equation in to form which is free from radicals.

\({\left[ {{{\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)}^{\frac{1}{2}}}} \right]^6} = {\left[ {{{\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]}^{\frac{1}{3}}}} \right]^6}\)

\(\Rightarrow {\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^3} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^2}\)

Now the differential equation is free from radicals

Degree of highest derivative = 3

Order and degree of the given differential equation = 4 and 3

Concept:

• Linear Differential Equation:

A general linear differential equation of order n, in the dependent variable y and the independent variable x, can be expressed in the following form:

\({{\rm{a}}_0}\left( {\rm{x}} \right)\frac{{{{\rm{d}}^{\rm{n}}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}} + {{\rm{a}}_1}\left( {\rm{x}} \right)\frac{{{{\rm{d}}^{{\rm{n}} - 1}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{{\rm{n}} - 1}}}} + \ldots + {{\rm{a}}_{{\rm{n}} - 1}}\left( {\rm{x}} \right)\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{a}}_{\rm{n}}}\left( {\rm{x}} \right){\rm{y}} = {\rm{b}}\left( {\rm{x}} \right)\)

A differential equation will be linear as long as,

1. The dependent variables and its derivatives occur in the first degree
2. The dependent variables and its derivatives are not multiplied together
3. There is no transcendental function (trigonometric, logarithmic, exponential, etc. of the dependent variable

 

• Applying this concept, it is evident that EQ 1 and EQ 3 are linear differential equations.
• EQ 2 is not a linear differential equation because the derivative of the dependent variable \(\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)\) occurs in 2^nd degree.
• EQ 4 is not a linear differential equation because it contains transcendental function of dependent variable (if any).

Concept:

If a first order linear differential equation can be expressed as \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{Py}} = {\rm{Q\;}}\)where P and Q are either functions of x or constants, then integrating factor (I.F.) is given by \({{\rm{e}}^{\smallint {\rm{Pdx}}}}\) and the solution of the differential equation is given by, \({\rm{y}}{{\rm{e}}^{\smallint {\rm{Pdx}}}} = \smallint {\rm{Q}} \times {{\rm{e}}^{\smallint {\rm{Pdx}}}}{\rm{dx}} + {\rm{c}}\) where c is an integrating constant.

Calculation:

The given differential equation \(\cos {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}}\tan {\rm{x}} = {\cos ^3}{\rm{x}}\) can be rearranged as \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + \frac{{{\rm{y}}\tan {\rm{x}}}}{{\cos {\rm{x}}}} = {\cos ^2}{\rm{x}}\)

\(\therefore \frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}}\tan {\rm{x}}\sec {\rm{x}} = {\cos ^2}{\rm{x\;}}\)

Hence the,

\({{\;I}}.{\rm{F}}. = {{\rm{e}}^{\smallint \tan {\rm{x}}\sec {\rm{xdx}}}} = {{\rm{e}}^{\sec {\rm{x}}}}\)

Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Roots of Auxiliary Equation	Complementary Function
m1, m2, m3, … (real and different roots)	\({C_1}{e^{{m_1}x}} + {C_2}{e^{{m_2}x}} + {C_3}{e^{{m_3}x}} + \ldots\)
m1, m1, m3, … (two real and equal roots)	\(\left( {{C_1} + {C_2}x} \right){e^{{m_1}x}} + {C_3}{e^{{m_3}x}} + \ldots\)
m1, m1, m1, m4… (three real and equal roots)	\(\left( {{C_1} + {C_2}x + {C_3}{x^2}} \right){e^{{m_1}x}} + {C_4}{e^{{m_4}x}} + \ldots\)
α + i β, α – i β, m3, … (a pair of imaginary roots)	\({e^{\alpha x}}\left( {{C_1}\cos \beta x + {C_2}\sin \beta x} \right) + {C_3}{e^{{m_3}x}} + \ldots\)
α ± i β, α ± i β, m5, … (two pairs of equal imaginary roots)	\({e^{\alpha x}}\left( {\left( {{C_1} + {C_2}x} \right)\cos \beta x + \left( {{C_3} + {C_4}x} \right)\sin \beta x} \right) + {C_5}{e^{{m_5}x}} + \ldots\)

 

Calculation:

\(\frac{{{d^2}y}}{{d{x^2}}} + a\frac{{dy}}{{dx}} + by = 0\)

\(\Rightarrow \left( {{D^2} + aD + b} \right)y = 0\)

Auxiliary equation:

D² + aD + b = 0

Roots of auxiliary equation are:

\(D = \frac{{ - a \pm \sqrt {{a^2} - 4b} }}{2}\)

Given that, roots of the auxiliary equation are real and equal.

⇒ D = -a/2

The general solution of the differential equation is: \(y = \left( {{c_1} + {c_2}x} \right){e^{ - \frac{{ax}}{2}}}\)

\(\left( {{\rm{x}}{{\rm{y}}^2} - {{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}}} \right){\rm{dx}} - {{\rm{x}}^2}{\rm{y\;dy}} = 0{\rm{\;}}\)

The above equations is in the form of

Mdx + N dy = 0

\(\begin{array}{l} M = x{y^2} - {{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}}\;\\ \end{array}\)

N = - x²y

\(\frac{{\partial M}}{{\partial y}} \ne \frac{{\partial N}}{{\partial x}}\)

Hence it is not an exact equation. So we have to reduce this equation to exact equation. For that we need to calculate integrating factor.

\(\frac{{\frac{{\partial {\rm{M}}}}{{\partial {\rm{y}}}} - \frac{{\partial {\rm{N}}}}{{\partial {\rm{x}}}}}}{{\rm{N}}} = {\rm{\;}}\frac{{2{\rm{xy}} - \left( { - 2{\rm{xy}}} \right)}}{{ - {{\rm{x}}^2}{\rm{y}}}} = \frac{{ - 4}}{{\rm{x}}}\)

This is the function of ‘x’ only.

\(NowI.F. = {e^{\smallint \frac{{ - 4}}{x}dx}} = {e^{ - 4\;logx\;}} = \frac{1}{{{x^4}}}\)

Multiplying the given equation by I.F. we can reduce in to exact equation.

\(\begin{array}{l} \Rightarrow \frac{1}{{{x^4}}}\left( {x{y^2} - {{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}}} \right)dx -\frac{1}{x^4} {x^2}y\;dy = 0\\ \Rightarrow \left( {\frac{{{y^2}}}{{{x^3}}} - \frac{1}{{{x^4}}}{{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}}} \right)dx - \frac{y}{{{x^2}}}dy = 0 \end{array}\)

Now the solution is,

\(\int_{y\;constant} {Mdx + \smallint \left( {{\rm{terms\;of\;N\;not\;Containing\;x}}} \right)dy = c}\)

\(\begin{array}{l} \Rightarrow \smallint \left( {\frac{{{{\rm{y}}^2}}}{{{{\rm{x}}^3}}} - \frac{1}{{{{\rm{x}}^4}}}{\rm{\;}}{{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}}} \right){\rm{dx}} = {\rm{C}}\\ \Rightarrow \frac{{ - {{\rm{y}}^2}}}{2}{{\rm{x}}^{-2}} + \frac{1}{3}\smallint {{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}}{\rm{\;}}\left( { - 3{{\rm{x}}^{ - 4}}} \right){\rm{dx}} = {\rm{C}}\\ \Rightarrow \frac{{ - {{\rm{y}}^2}}}{2}{{\rm{x}}^{-2}} + \frac{1}{3}{\rm{\;}}{{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}} = {\rm{C}}\\ \Rightarrow \frac{1}{3}{\rm{\;}}{{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}} - \frac{{{{\rm{y}}^2}{{\rm{x}}^{-2}}}}{2} = {\rm{C\;}} \Rightarrow 2{{\rm{e}}^{\frac{1}{{{{\rm{x}}^3}}}}} - 3{{\rm{y}}^2}{{\rm{x}}^{-2}} = {\rm{C}} \end{array}\)

Concept:

Rules for finding Particular Integral (P.I.) for exponential function:

Particular integral of \(\left\{ {{\rm{f}}\left( {\rm{D}} \right)} \right\}{\rm{y}} = {\rm{\;}}{{\rm{e}}^{{\rm{ax}}}}{\rm{\;}}\) is calculated by,

\(\frac{1}{{{\rm{f}}\left( {\rm{D}} \right)}}{{\rm{e}}^{{\rm{ax}}}} = \frac{1}{{{\rm{f}}\left( {\rm{a}} \right)}}{{\rm{e}}^{{\rm{ax}}}}{\rm{\;\;provided\;f}}\left( {\rm{a}} \right) \ne 0{\rm{\;\;}}\)

\({\rm{If\;f}}\left( {\rm{a}} \right) = 0,{\rm{\;then\;P}}.{\rm{I}}.{\rm{\;is\;calculated\;as\;}}\frac{1}{{{\rm{f}}\left( {\rm{D}} \right)}}{{\rm{e}}^{{\rm{ax}}}} = {\rm{x}}\frac{1}{{{\rm{f'}}\left( {\rm{a}} \right)}}{{\rm{e}}^{{\rm{ax}}}}{\rm{\;\;\;provided\;f'}}\left( {\rm{a}} \right) \ne 0{\rm{\;\;}}\)

\({\rm{If\;f'}}\left( {\rm{a}} \right) = 0,{\rm{\;then\;P}}.{\rm{I}}.{\rm{\;is\;calculated\;as\;}}\frac{1}{{{\rm{f}}\left( {\rm{D}} \right)}}{{\rm{e}}^{{\rm{ax}}}} = {{\rm{x}}^2}\frac{1}{{{\rm{f''}}\left( {\rm{a}} \right)}}{{\rm{e}}^{{\rm{ax}}}}{\rm{\;\;\;provided\;f''}}\left( {\rm{a}} \right) \ne 0{\rm{\;and\;so\;on}}.{\rm{\;\;}}\)

Calculation:

\(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} - \frac{{{\rm{dy}}}}{{{\rm{dx}}}} - 20{\rm{y}} = {\rm{\;\;}}\sinh 5{\rm{x}} = \frac{1}{2}\left( {{{\rm{e}}^{5{\rm{x}}}} - {{\rm{e}}^{ - 5{\rm{x}}}}} \right){\rm{\;}}\therefore {\rm{\;}}\left( {{{\rm{D}}^2} - {\rm{D}} - 20} \right){\rm{y}} = \frac{1}{2}\left( {{{\rm{e}}^{5{\rm{x}}}} - {{\rm{e}}^{ - 5{\rm{x}}}}} \right)\)

\({\rm{Here}},{\rm{\;f}}\left( {\rm{D}} \right) = {{\rm{D}}^2} - {\rm{D}} - 20{\rm{\;}}\)

\(\therefore {\rm{P}}.{\rm{I}}.{\rm{\;}} = \frac{1}{{{{\rm{D}}^2} - {\rm{D}} - 20}} \times \frac{1}{2}\left( {{{\rm{e}}^{5{\rm{x}}}} - {{\rm{e}}^{ - 5{\rm{x}}}}} \right) = \frac{1}{2}\left[ {\frac{1}{{{{\rm{D}}^2} - {\rm{D}} - 20}}{{\rm{e}}^{5{\rm{x}}}} - \frac{1}{{{{\rm{D}}^2} - {\rm{D}} - 20}}{{\rm{e}}^{ - 5{\rm{x}}}}} \right]\)

Now, \({\rm{f}}\left( 5 \right) = 0{\rm{\;and\;\;f}}\left( { - 5} \right) \ne 0\) Hence, approaching with proper rule we get,

\({\rm{P}}.{\rm{I}}. = \frac{1}{2}\left[ {{\rm{x}}\frac{1}{{2{\rm{D}} - 1}}{{\rm{e}}^{5{\rm{x}}}} - \frac{1}{{{{\left( { - 5} \right)}^2} - \left( { - 5} \right) - 20}}{{\rm{e}}^{ - 5{\rm{x}}}}} \right] = \frac{1}{2}\left[ {{\rm{x}}\frac{1}{{2 \times 5 - 1}}{{\rm{e}}^{5{\rm{x}}}} - \frac{1}{{10}}{{\rm{e}}^{ - 5{\rm{x}}}}} \right] = \frac{1}{2}\left[ {\frac{{\rm{x}}}{9}{{\rm{e}}^{5{\rm{x}}}} - \frac{1}{{10}}{{\rm{e}}^{ - 5{\rm{x}}}}} \right]\)

Mistake Point:

Notice the hyperbolic function \(\sinh 5{\rm{x}} = \frac{1}{2}\left( {{{\rm{e}}^{5{\rm{x}}}} - {{\rm{e}}^{ - 5{\rm{x}}}}} \right)\).

The rules are completely different for trigonometric functions (such as \(\sin 5{\rm{x}},{\rm{\;}}\cos 5{\rm{x}})\) on hyperbolic functions (such as \(\sinh 5{\rm{x}},{\rm{\;}}\cosh 5{\rm{x}})\). 

Given differential equation is,

dx – (x + y + 1) dy = 0

\(\frac{{dy}}{{dx}} = \frac{1}{{\left( {x + y + 1} \right)}}\) 

Put (x + y + 1) = t

\( \Rightarrow 1 + \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}\) 

\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} - 1\) 

Now, the differential equation becomes

\(\frac{{dt}}{{dx}} - 1 = \frac{1}{t}\) 

\(\frac{{dt}}{{dx}} = \frac{1}{t} + 1\) 

\(\frac{{dt}}{{dx}} = \frac{{t + 1}}{t}\) 

\( \Rightarrow dt\left( {\frac{t}{{t + 1}}} \right) = dx\) 

\( \Rightarrow \left( {1 - \frac{1}{{t + 1}}} \right)dt = dx\) 

\( \Rightarrow \smallint \left( {1 - \frac{1}{{t + 1}}} \right)dt = dx\) 

⇒ t = In (t + 1) = x + C₁

⇒ (x + y + 1) – In (x + y + 2) = x + C₁

⇒ y + 1 – C₁ = In (x + y + 2)

⇒ In (x + y + 2) = y + k

⇒ (x + y + 2) = e^y - e^k

⇒ (x + y + 2) e^-y = C

Given equation is a linear differential equation.

D² + 2D + 17 = 0

\({\rm{D}} = \frac{{ - 2 \pm \sqrt {4 - 4 \times 17 \times 1} }}{{2 \times 1}}\)

D = -1 ± 4i

y = C₁e^{(-1 + 4i)x} + C₂e^{(-1 - 4i)x}

y = e^–x[C₁e^4xi + C₂e^–4xi]

y = e^–x[C₁(Cos4x + iSin4x) + C₂[Cos(-4x) + iSin(-4x)]]

y = e^–x[(C₁ + C₂)Cos4x + (C₁ + C₂) iSin4x]

Let C₁ + C₂ = C₃

(C₁ - C₂)i = C₄

y = e^–x[C₃Cos4x + C₃Sin4x]

For x = 0, y = 2

2 = C₃

For x = \(\frac{\pi }{2}\), y’ = 0

\(\frac{{dy}}{{dx}} = {{\rm{e}}^{ - x}}\left[ { - 4{{\rm{C}}_3}{\rm{Sin}}4{\rm{x}} + 4{{\rm{C}}_4}{\rm{Cos}}4{\rm{x}}} \right] + \left[ {{{\rm{C}}_3}{\rm{Cos}}4{\rm{x}} + {{\rm{C}}_4}{\rm{Sin}}4{\rm{x}}} \right]{{\rm{e}}^{ - x}}\left( { - 1} \right)\)

\(0 = {{\rm{e}}^{ - \frac{\pi }{2}}}\left[ { - 4{{\rm{C}}_3}{\rm{Sin}}2{\rm{\pi \;}} + 4{{\rm{C}}_4}{\rm{Cos}}2{\rm{\pi }}} \right] - {{\rm{e}}^{ - \frac{\pi }{2}}}\left[ {{{\rm{C}}_3} + {{\rm{C}}_4}{\rm{Sin}}2{\rm{\pi }}} \right]\)

\(0 = {{\rm{e}}^{ - \frac{\pi }{2}}}\left[ {4{{\rm{C}}_4}} \right] - {{\rm{e}}^{ - \frac{\pi }{2}}}\left[ {{{\rm{C}}_3}} \right]\)

\(0 = {{\rm{e}}^{ - \frac{\pi }{2}}}\left[ {4{{\rm{C}}_4} - {{\rm{C}}_3}} \right]\)

\(4{{\rm{C}}_4} - {{\rm{C}}_3} = 0\)

For C₃ = 2

4C₄ - 2 = 0

\({\rm{C}}4 = \frac{2}{4} = \frac{1}{2}\)

C₃ = 2

\({\rm{y}} = {{\rm{e}}^{ - {\rm{x}}}}\left[ {2{\rm{Cos}}4{\rm{x}} + \frac{1}{2}{\rm{Sin}}4{\rm{x}}} \right]\)

Concept:

Cauchy-Euler Equation:

The homogeneous linear differential equations in form of  \({{\rm{x}}^{\rm{n}}}\frac{{{{\rm{d}}^{\rm{n}}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}} + {{\rm{a}}_1}{{\rm{x}}^{{\rm{n}} - 1}}\frac{{{{\rm{d}}^{{\rm{n}} - 1}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{{\rm{n}} - 1}}}} + \ldots + {{\rm{a}}_{{\rm{n}} - 1}}{\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{a}}_{\rm{n}}}{\rm{y}} = {\rm{b}}\left( {\rm{x}} \right)\) is known as Cauchy-Euler Equation, where  are constants and b(x) is either a function of x or a constant. The solution of these equations can be easily obtained by substituting \({\rm{x}} = {{\rm{e}}^{\rm{t}}}\) and this will give the following results:

\({\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{Dy}},{\rm{\;\;\;\;}}{{\rm{x}}^2}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{D}}\left( {{\rm{D}} - 1} \right){\rm{y}},{\rm{\;\;}}{{\rm{x}}^3}\frac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}} = {\rm{D}}\left( {{\rm{D}} - 1} \right)\left( {{\rm{D}} - 2} \right),{\rm{\;\;}} \ldots ,{\rm{\;\;}}{{\rm{x}}^{\rm{n}}}\frac{{{{\rm{d}}^{\rm{n}}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}} = {\rm{D}}\left( {{\rm{D}} - 1} \right) \ldots \left( {{\rm{D}} - \left( {{\rm{n}} - 1} \right)} \right){\rm{y}}\)

where D =d/dt operator.

Rules for finding Particular Integral (P.I.) for exponential function:

Particular integral of \(\left\{ {{\rm{f}}\left( {\rm{D}} \right)} \right\}{\rm{y}} = {\rm{\;}}{{\rm{e}}^{{\rm{ax}}}}{\rm{\;}}\) is calculated by,

\(\frac{1}{{{\rm{f}}\left( {\rm{D}} \right)}}{{\rm{e}}^{{\rm{ax}}}} = \frac{1}{{{\rm{f}}\left( {\rm{a}} \right)}}{{\rm{e}}^{{\rm{ax}}}}{\rm{\;\;provided\;f}}\left( {\rm{a}} \right) \ne 0{\rm{\;\;}}\)

\({\rm{If\;f}}\left( {\rm{a}} \right) = 0,{\rm{\;then\;P}}.{\rm{I}}.{\rm{\;is\;calculated\;as\;}}\frac{1}{{{\rm{f}}\left( {\rm{D}} \right)}}{{\rm{e}}^{{\rm{ax}}}} = {\rm{x}}\frac{1}{{{\rm{f'}}\left( {\rm{a}} \right)}}{{\rm{e}}^{{\rm{ax}}}}{\rm{\;\;\;provided\;f'}}\left( {\rm{a}} \right) \ne 0{\rm{\;and\;so\;on}}.{\rm{\;\;}}\)

Calculation:

The given equation can be rearranged as, \({{\rm{x}}^2}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} + 2{\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - 6{\rm{y\;}} = 7{{\rm{x}}^4}\). So, it is in the form of Cauchy-Euler Equation. So, by substituting \({\rm{x}} = {{\rm{e}}^{\rm{t}}}\) we get,

\(\left[ {{\rm{D}}\left( {{\rm{D}} - 1} \right) + 2{\rm{D}} - 6} \right]{\rm{y}} = 7{{\rm{e}}^{4{\rm{t}}}}{\rm{\;}} \Rightarrow \left( {{{\rm{D}}^2} + {\rm{D}} - 6} \right){\rm{y}} = 7{{\rm{e}}^{4{\rm{t}}}}{\rm{\;\;where\;D}} = \frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{operator}}\)

So, the auxiliary equation is, \({{\rm{D}}^2} + {\rm{D}} - 6 = 0 \Rightarrow {\rm{D}} = - 3,{\rm{\;}}2\)

\(\therefore {\rm{\;The\;Complementary\;function}}\left( {{\rm{C}}.{\rm{F}}.} \right) = {\rm{\;A}}{{\rm{e}}^{2{\rm{t}}}} + {\rm{B}}{{\rm{e}}^{ - 3{\rm{t}}}}\)

\(\therefore {\rm{\;The\;Particular\;integral\;}}\left( {{\rm{P}}.{\rm{I}}.} \right) = \frac{1}{{{{\rm{D}}^2} + {\rm{D}} - 6}}7{{\rm{e}}^{4{\rm{t}}}} = 7 \times \frac{1}{{{4^2} + 4 - 6}}{{\rm{e}}^{4{\rm{t}}}} = \frac{{{{\rm{e}}^{4{\rm{t}}}}}}{2}\)

\(\therefore {\rm{\;The\;total\;solution}},{\rm{y}} = {\rm{C}}.{\rm{F}}.{\rm{\;}} + {\rm{P}}.{\rm{I}}.{\rm{\;}} = {\rm{\;A}}{{\rm{e}}^{2{\rm{t}}}} + {\rm{B}}{{\rm{e}}^{ - 3{\rm{t}}}} + \frac{{{{\rm{e}}^{4{\rm{t}}}}}}{2}{\rm{\;}} \Rightarrow {\rm{y}} = {\rm{A}}{{\rm{x}}^2} + \frac{{\rm{B}}}{{{{\rm{x}}^3}}} + \frac{1}{2}{{\rm{x}}^4}\)

\({\rm{Given}},{\rm{\;y\;}} = {\rm{\;}}0{\rm{\;\;at\;x\;}} = {\rm{\;}}1{\rm{\;}}\therefore 0 = {\rm{A}} \times {1^2} + \frac{{\rm{B}}}{{{\rm{\;}}{1^3}}} + \frac{1}{2} \Rightarrow {\rm{A}} + {\rm{B}} = {\rm{\;}} - 0.5\)

\({\rm{Given}},{\rm{\;y\;}} = {\rm{\;}}1{\rm{\;\;at\;x\;}} = {\rm{\;}} - 1{\rm{\;}}\therefore 1 = {\rm{A}} \times {\left( { - 1} \right)^2} + \frac{{\rm{B}}}{{{{\left( { - {\rm{\;}}1} \right)}^3}}} + \frac{1}{2} \times 1 \Rightarrow {\rm{A}} - {\rm{B}} = {\rm{\;}}0.5{\rm{\;\;\;\;}}\therefore {\rm{A}} = 0{\rm{\;}}\& \;\;B = - 0.5\)  

\(\therefore {\rm{y}}\left( {\rm{x}} \right) = \frac{1}{2}\left( {{{\rm{x}}^4} - \frac{1}{{{{\rm{x}}^3}}}} \right){\rm{\;\;\;}}\therefore {\rm{\;y}}\left( 2 \right) = \frac{1}{2}\left( {{2^4} - \frac{1}{{{2^3}}}} \right) = 7.9375\)

Concept:

The most convenient way to solve a first-order linear differential equation is by implementing the “integrating factor approach”. To apply the “integrating factor approach”, the equation must be expressed as \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{Py}} = {\rm{Q\;}}\), where P and Q are either function of x or constants. Then the integrating factor (I.F.) can be given by \({{\rm{e}}^{\smallint {\rm{Pdx}}}}\) and the solution of the differential equation is given by, \({\rm{y}}{{\rm{e}}^{\smallint {\rm{Pdx}}}} = \smallint {\rm{Q}} \times {{\rm{e}}^{\smallint {\rm{Pdx}}}}{\rm{dx}} + {\rm{c}}\) where c is an integrating constant.

If a linear first-order differential equation is not given in form of \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{Py}} = {\rm{Q}}\), then if possible, it’s better to bring it to that form either by substituting or rearranging.

Calculation:

Given the differential equation, \({\sec ^2}{\rm{y}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{x}}^2}\tan {\rm{y}} = {{\rm{x}}^2}\) which is not in the \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{Py}} = {\rm{Q}}\) form.

Put, \(\tan {\rm{y}} = {\rm{t\;\;\;}}\therefore {\rm{\;}}\frac{{{\rm{dt}}}}{{{\rm{dx}}}} = {\sec ^2}{\rm{y}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\)

Hence the equation reduces to, \(\frac{{{\rm{dt}}}}{{{\rm{dx}}}} + {{\rm{x}}^2}{\rm{t}} = {{\rm{x}}^2}\) which is now in the \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{Py}} = {\rm{Q}}\) form.

\(\therefore {\rm{\;I}}.{\rm{F}}. = {{\rm{e}}^{\smallint {{\rm{x}}^2}{\rm{dx}}}} = {{\rm{e}}^{\frac{{{{\rm{x}}^3}}}{3}}}\)

Hence, the solution is \(,{\rm{\;\;t}} \times {{\rm{e}}^{\frac{{{{\rm{x}}^3}}}{3}}} = \smallint {{\rm{x}}^2} \times {{\rm{e}}^{\frac{{{{\rm{x}}^3}}}{3}}}{\rm{dx}} + {\rm{c}} = \smallint {{\rm{e}}^{\frac{{{{\rm{x}}^3}}}{3}}}{\rm{d}}\left( {\frac{{{{\rm{x}}^3}}}{3}} \right) + {\rm{c}} = {{\rm{e}}^{\frac{{{{\rm{x}}^3}}}{3}}} + {\rm{c}}\)

\(\therefore {\rm{t}} = 1 + {\rm{c}}{{\rm{e}}^{ - \frac{{{{\rm{x}}^3}}}{3}}}\)  Hence, by replacing t we get, \(\tan {\rm{y}} = {\rm{\;}}1 + {\rm{c}}{{\rm{e}}^{ - \frac{{{{\rm{x}}^3}}}{3}}}{\rm{\;}}\)

Given, \({\rm{y}}\left( 0 \right) = 2.5{\rm{\;\;\;\;\;}}\therefore \tan 2.5 = {\rm{\;}}1 + {\rm{c}}{{\rm{e}}^{ - {\rm{\;}}\frac{{{0^3}}}{3}}}{\rm{\;}} \Rightarrow {\rm{c}} = {\rm{\;}} - 1.747{\rm{\;\;}}\)

The final form of the solution is, \(\tan {\rm{y}} = {\rm{\;}}1 - 1.747{{\rm{e}}^{ - {\rm{\;}}\frac{{{{\rm{x}}^3}}}{3}}}\)

\(\therefore {\rm{at\;x}} = 1,{\rm{\;}}\tan {\rm{y}} = {\rm{\;}}1 - 1.747{{\rm{e}}^{ - {\rm{\;}}\frac{{{1^3}}}{3}}} = - 0.2518{\rm{\;\;}} \Rightarrow {\rm{y}} = - 0.2466\)

Mistake Point:

Make sure to change the argument of trigonometric function into the “radian” of “degree” in the calculator. In general, by default, the argument of a trigonometric function is set to “degree” in online Calculator.