Mathematics Test - 20

Concept:

A differential equation is said to be linear when

a) Dependent variable and its derivative should have power ‘1’

b) Dependent variable and its derivatives can have a product with independent variable

c) Dependent variable and its derivatives can’t have product

Explanation:

In the given differential equations, y is the dependent variable and x is the independent variable.

We know that,

The linear differential equation should not have terms like \(y\frac{dy}{dx}\) .

Hence,

\(y\frac{dy}{dx}+4x=0\)

is a non-linear differential equation.

Concept:

In first order linear differential equation;

\(\rm {dy\over dx}+Py=Q\), where P and Q are function of x

Integrating factor (IF) = e∫ P dx

y × (IF) = ∫ Q(IF) dx

Calculation:

\(\rm {dy\over dx}+ y \sec^2 x = 3\cos x\)

IF = e∫ sec² x dx

⇒ IF = e^tan x

Concept:

• Order: The order of a differential equation is the order of the highest derivative appearing in it.
• Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

For example, if the equation is

\({\left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}} \right)^3} + \frac{{{\rm{dy}}}}{{{\rm{dx}}}} - {{\rm{x}}^2} = 0\)

Here, order of the highest derivative is 2. And the degree is 3.

Calculation:

Given: \({\left( {\frac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}}} \right)^2} = {{\rm{y}}^4} + {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^5}\)

To find: Order and degree

So, the highest order is \(\frac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}}\)

∴ Order is 3.

Degree of differential equation is the degree highest order. So, degree is 2.

Calculation:

\(\begin{array}{l} \rm \Rightarrow \frac{d y}{d x}=e^{x} e^{y}+x^{2} e^{y} \\ \rm \Rightarrow \frac{d y}{d x}=e^{y}\left(e^{x}+x^{2}\right) \\ \rm \Rightarrow \frac{d y}{e^{y}}=\left(e^{x}+x^{2}\right) d x \\\Rightarrow \int \frac{d y}{e^{y}}=\int e^{x} d x+\int x^{2} d x+c_{1} \\ \rm \Rightarrow \int e^{-y} d y=e^{x}+\frac{1}{3} x^{3}+c_1 \\ \rm \Rightarrow -e^{-y}=e^{x}+\frac{1}{3} x^{3}+c_1 \end{array}\\\)

\(\rm \Rightarrow e^{x} + e^{-y} + \frac {x^3} 3 = c\)

Hence, option (2) is correct.

Concept:

First Order Linear Differential Equation:
A differential equation of the from \(\rm \dfrac{dy}{dx}\) + P × y = Q, where P and Q are constants or functions of x only, is known as a first order linear differential equation.

Steps to solve a First Order Linear Differential Equation:

• Convert into the standard form \(\rm \dfrac{dy}{dx}\) + P × y = Q, where P and Q are constants or functions of x only.
• Find the Integrating Factor (F) by using the formula: F = \(\rm e^{\int P\ dx}\).
• Write the solution using the formula: \(\rm y\times F=\displaystyle \int (Q\times F)\ dx+C\) where C is the constant of integration.

Calculation:

\(\rm x \dfrac{dy}{dx}-y=3\)

\(\rm \Rightarrow \dfrac{dy}{dx}+\left(\dfrac{-1}{x}\right)y=\dfrac{3}{x}\)

⇒ P = \(\rm \dfrac{-1}{x}\) and Q = \(\rm \dfrac{3}{x}\).

Integrating factor F = \(\rm e^{\int P\ dx}=e^{\int \tfrac{-1}{x}\ dx}=e^{-\ln x}=\dfrac{1}{x}\).

The solution of the given differential equation is:

\(\rm y\times \dfrac{1}{x}=\displaystyle \int \left(\dfrac{3}{x}\times \dfrac{1}{x}\right)\ dx+C\)

\(\rm \Rightarrow \dfrac{y}{x}=3\left(\dfrac{-1}{x}\right)+C \)

⇒ y = Cx - 3, which is the equation of a straight line.

Concept:

To find the solution of given differential equation, convert the given differential equation in variable separable form and then integrate it.

Calculations:

Given, \(\rm \frac {dy}{dx} \cos (x - y) = 1\)

⇒ \(\rm \dfrac {dx}{dy} = \cos (x - y)\)....(1)

put x  - y = u

Differentiate w. r. to y, we get

⇒ \(\rm \dfrac {dx}{dy} - 1 = \dfrac {du}{dy}\)

⇒ \(\rm \dfrac {dx}{dy} = 1 + \dfrac {du}{dy}\)

Equation (1) becomes,

⇒\(\rm 1 + \dfrac {du}{dy} = cos \;u\)

⇒\(\rm \dfrac {du}{dy} = cos \;u -1\)

⇒\(\rm \dfrac {du}{cos \;u -1} = dy\)

variables are separated. Integrating on both side, we get

⇒\(\rm \int \dfrac {du}{cos \;u -1} =\int dy\)

⇒\(\rm \int \dfrac {du}{-2\;sin^2 \dfrac u 2} =\int dy\)

⇒\(\rm -\dfrac 12 \int cosec^2 (\dfrac u 2)\; du =\int dy\)

\(\rm ⇒\cot \left( \frac {u}{2} \right) = y + c\)

​\(\cot \left( \frac {x - y}{2} \right) = y + c\)​

Concept:

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

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

Calculation:

The differential equation is given as: \(\rm \frac{d^3y}{dx^3} + \cos\left(\frac{d^2y}{dx^2}\right) = 0\)

The highest order derivative presents in the differential equation is \(\rm \frac{d^3y}{dx^3}\)

Hence, its order is three.

Here the given differential equation is not a polynomial equation, Hence its degree is not defined.

Calculation:

Given: A particle moves along a straight line with velocity given by \(\frac{{dy}}{{dt}} = \left( {1 + y} \right)\) where ‘y’ is distance travelled

So, in order to find the time taken by a particle to travel distance of 999 meter we have to integrate the above equation

By variable separable method the differential equation \(\frac{{dy}}{{dt}} = \left( {1 + y} \right)\) can re-written as: \(\frac{{dy}}{{\left( {1 + y} \right)}} = dt\)

Integrating both sides and taking limits we get

\(\mathop \smallint \limits_0^{999} \frac{{dy}}{{\left( {1 + y} \right)}} = \mathop \smallint \limits_0^t dt\)

\(⇒ \left[ {ln\;\left( {1 + y} \right)} \right]_0^{999} = t\)

⇒ In (1000) – In (1) = t

⇒ t = 3 log_e 10

Hence, option 3 is the correct answer.

Concept:

• Equation of circle having centre (h, k) and radius r is (x - h) ² + (y - k) ² = r²

Calculation:

Let equation of family of circles passing through origin and having centre (h, 0) be

(x - h) ² + (y - 0) ² = h²

⇒ x² + h² - 2xh + y² = h²

⇒ x² - 2xh + y² = 0      ---(1)

Differentiating both sides, we get

⇒ 2x - 2h + 2y(dy/dx) = 0

⇒ x + y(dy/dx) - h = 0      ---(2)

From equation 1^st,

x² - 2xh + y² = 0

\(\therefore {\rm{h}} = \frac{{{{\rm{x}}^2} + {{\rm{y}}^2}}}{{2{\rm{x}}}}\)

Put the value of h in equation 2^nd,

\(\Rightarrow {\rm{x}} + {\rm{y}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \left( {\frac{{{{\rm{x}}^2} + {{\rm{y}}^2}}}{{2{\rm{x}}}}} \right) = 0{\rm{\;}}\)

\(\Rightarrow 2{{\rm{x}}^2} + 2{\rm{xy}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - {{\rm{x}}^2} - {{\rm{y}}^2} = 0\)

\(\Rightarrow 2{\rm{xy\;}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {{\rm{y}}^2} - {{\rm{x}}^2}\)

Hence option 2^nd is correct.

Concept:

• \(\smallint \frac{{{\rm{dx}}}}{{\sqrt {\left( {{{\rm{a}}^2} - {{\rm{x}}^2}} \right)} }} = {\rm{\;}}{\sin ^{ - 1}}\left( {\frac{{\rm{x}}}{{\rm{a}}}} \right) + {\rm{c}}\)
• \(\smallint \sqrt {\left( {{{\rm{a}}^2} - {{\rm{x}}^2}} \right)} {\rm{\;dx}} = {\rm{\;}}\frac{{\rm{x}}}{2}\sqrt {\left( {{{\rm{a}}^2} - {{\rm{x}}^2}} \right)} + {\rm{\;}}\frac{{{{\rm{a}}^2}}}{2}{\sin ^{ - 1}}(\frac{x}{a}) + {\rm{c}}\)

Calculation:

Given:

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \sqrt {1 - {{\rm{x}}^2} - {{\rm{y}}^2} + {{\rm{x}}^2}{{\rm{y}}^2}} \)

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \sqrt {\left( {1 - {{\rm{x}}^2}} \right) - {{\rm{y}}^2}\left( {1 - {{\rm{x}}^2}} \right)} = \sqrt {\left( {1 - {{\rm{x}}^2}} \right)\left( {1 - {{\rm{y}}^2}} \right)} \)

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \sqrt {\left( {1 - {{\rm{x}}^2}} \right)} {\rm{\;}} \times {\rm{\;}}\sqrt {\left( {1 - {{\rm{y}}^2}} \right)} \)

\(\Rightarrow \frac{{{\rm{dy}}}}{{\sqrt {\left( {1 - {{\rm{y}}^2}} \right)} }} = \sqrt {\left( {1 - {{\rm{x}}^2}} \right)} {\rm{\;dx\;}}\)

On integrating both sides, we get

\(\Rightarrow \smallint \frac{{{\rm{dy}}}}{{\sqrt {\left( {1 - {{\rm{y}}^2}} \right)} }} = \smallint \sqrt {\left( {1 - {{\rm{x}}^2}} \right)} {\rm{\;dx}}\)

\(\Rightarrow {\rm{\;}}{\sin ^{ - 1}}{\rm{y}} = {\rm{\;}}\frac{{\rm{x}}}{2}\sqrt {\left( {1 - {{\rm{x}}^2}} \right)} + {\rm{\;}}\frac{1}{2}{\sin ^{ - 1}}{\rm{x}} + {\rm{c}}\)

\(2{\sin ^{ - 1}}{\rm{y}} = {\rm{x}}\sqrt {1 - {{\rm{x}}^2}} + {\sin ^{ - 1}}{\rm{x}} + {\rm{c}}\)