12th NCERT Differential Equation Exercise Miscellaneous Number of questions 18
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Question (1)

For each of the fifferential equations given below, indicate its order and degree ( if defined)
$(i)\frac{{{d^2}y}}{{d{x^2}}} + 5x{\left( {\frac{{dy}}{{dx}}} \right)^2} - 6y = \log x$ $(ii){\left( {\frac{{dy}}{{dx}}} \right)^3} - 4{\left( {\frac{{dy}}{{dx}}} \right)^2} + 7y = \sin x$ $(iii)\frac{{{d^4}y}}{{d{x^4}}} - \sin \left( {\frac{{{d^3}y}}{{d{x^3}}}} \right) = 0$

Solution

$(i)\frac{{{d^2}y}}{{d{x^2}}} + 5x{\left( {\frac{{dy}}{{dx}}} \right)^2} - 6y = \log x$
Order = 2
Degree = 1
$(ii){\left( {\frac{{dy}}{{dx}}} \right)^3} - 4{\left( {\frac{{dy}}{{dx}}} \right)^2} + 7y = \sin x$
Order = 1
Degree = 3
$(iii)\frac{{{d^4}y}}{{d{x^4}}} - \sin \left( {\frac{{{d^3}y}}{{d{x^3}}}} \right) = 0$
Order = 4
Degree not defined as composite of derivative

Question (2)

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$(i)y = a{e^x} + b{e^{ - x}} + {x^2}\quad :x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2 = 0$
$(ii)y = {e^x}\left( {a\cos x + b\sin x} \right)\quad :\frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + 2y = 0$
$(iii)y = x\sin 3x \quad :\frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x = 0$ $(iv){x^2} = 2{y^2}\log y\quad :\left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy = 0$

Solution

$(i)y = a{e^x} + b{e^{ - x}} + {x^2}\quad :x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2 = 0$
Differentiate w.r.t. 'x' $x\frac{{dy}}{{dx}} + y = a{e^x} + b{e^{ - x}}\left( { - 1} \right) + 2x$ $x\frac{{dy}}{{dx}} + y = a{e^x} - b{e^{ - x}} + 2x$ Differentiate w.r.t. 'x' again $x\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}}\left( 1 \right) + \frac{{dy}}{{dx}} = a{e^x} - b{e^{ - x}}\left( { - 1} \right) + 2$ $x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} = a{e^x} + b{e^{ - x}} + 2$ $x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} = xy - {x^2} + 2$ $x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2 = 0$
$(ii)y = {e^x}\left( {a\cos x + b\sin x} \right)\quad :\frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + 2y = 0$
$y = {e^x}\left( {a\cos x + b\sin x} \right)$
Differentiate w.r.t. 'x'
$\frac{{dy}}{{dx}} = {e^x}\left[ {a\left( { - \sin x} \right) + b\cos x} \right] + \left( {a\cos x + b\sin x} \right){e^x}$ $\frac{{dy}}{{dx}} = {e^x}\left( { - a\sin x + b\cos x} \right) + y$ $\frac{{dy}}{{dx}} - y = {e^x}\left( { - a\sin x + b\cos x} \right)$ Differentiate w.r.t. 'x'
$\frac{{{d^2}y}}{{d{x^2}}} - \frac{{dy}}{{dx}} = {e^x}\left[ { - a\cos x + b\left( { - \sin x} \right)} \right] + \left( { - a\sin x + b\cos x} \right){e^x}$ $\frac{{{d^2}y}}{{d{x^2}}} - \frac{{dy}}{{dx}} = - {e^x}\left( {a\cos x + b\sin x} \right) + \left( {\frac{{dy}}{{dx}} - y} \right)$ $\frac{{{d^2}y}}{{d{x^2}}} - \frac{{dy}}{{dx}} = - y + \frac{{dy}}{{dx}} - y$ $\frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + 2y = 0$
$(iii)y = x\sin 3x \quad :\frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x = 0$ y= x sin3x
Differentiate w.r.t. 'x'
$\frac{{dy}}{{dx}} = x\cos 3x\left( 3 \right) + \sin 3x$ $\frac{{dy}}{{dx}} = 3x\cos 3x + \sin 3x$ Differentiate w.r.t. 'x'
$\frac{{{d^2}y}}{{d{x^2}}} = 3\left[ {x\left( { - \sin 3x} \right)3 + \cos 3x\left( 1 \right)} \right] + \cos 3x\left( x \right)$ $\frac{{{d^2}y}}{{d{x^2}}} = 3\left( { - 3x\sin 3x + \cos 3x} \right) + 3\cos 3x$ $\frac{{{d^2}y}}{{d{x^2}}} = - 9x\sin 3x + 3\cos 3x + 3\cos 3x$ $\frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x = 0$
$(iv){x^2} = 2{y^2}\log y\quad :\left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy = 0$
${x^2} = 2{y^2}\log y$ Differentiate w.r.t. 'x'
$\require{cancel} \cancel{2}x = \cancel{2}\left[ {{y^2}\frac{1}{y}\frac{{dy}}{{dx}} + \log y\left( {2y} \right)\frac{{dy}}{{dx}}} \right]$ $x = y\frac{{dy}}{{dx}} + 2y\log y\frac{{dy}}{{dx}}$ $x = \frac{{dy}}{{dx}}\left( {y + 2y\log y} \right)$ Multiply by y
$xy = \frac{{dy}}{{dx}}\left( {{y^2} + 2{y^2}\log y} \right)$ $xy = \frac{{dy}}{{dx}}\left( {{y^2} + {x^2}} \right)$ $\left( {{y^2} + {x^2}} \right)\frac{{dy}}{{dx}} - xy = 0$

Question (3)

From the differential equation representing the family of curves given by
${\left( {x - a} \right)^2} + 2{y^2} = {a^2}$ where a is an arbitrary constant

Solution

From the equation of family
${\left( {x - a} \right)^2} + 2{y^2} = {a^2}$ $\Rightarrow {x^2} - 2ax + {a^2} + 2{y^2} = {a^2}$ $\Rightarrow {x^2} - 2ax + 2{y^2} = 0$ Differentiate w.r.t. 'x'
$2\left( {x - a} \right) + 4y\frac{{dy}}{{dx}} = 0$ $x - a + 2y\frac{{dy}}{{dx}} = 0$ $x + 2y\frac{{dy}}{{dx}} = a$ Replacing value of 'a'
${x^2} - 2x\left( {x + 2y{y_1}} \right) + 2{y^2} = 0$ ${x^2} - 2{x^2} - 4xy{y_1} + 2{y^2} = 0$ $- {x^2} - 4xy{y_1} + 2{y^2} = 0$ $\therefore 2{y^2} - {x^2} = 4xy{y_1}$ $\therefore {y_1} = \frac{{2{y^2} - {x^2}}}{{4xy}}$

Question (4)

Prove that ${x^2} - {y^2} = c{\left( {{x^2} + {y^2}} \right)^2}$ is the general solution of differential quation
$\left( {{x^3} - 3x{y^2}} \right)dx = \left( {{y^3} - 3{x^2}y} \right)dy$, where c is a parameter

Solution

$\left( {{x^3} - 3x{y^2}} \right)dx = \left( {{y^3} - 3{x^2}y} \right)dy$ $\frac{{dy}}{{dx}} = \frac{{{x^3} - 3x{y^2}}}{{{y^3} - 3{x^2}y}}$ $\frac{{dy}}{{dx}} = \frac{{\cancel{{x^3}}\left( {1 - 3\frac{{{y^2}}}{{{x^2}}}} \right)}}{{\cancel{{x^3}}\left[ {{{\left( {\frac{y}{x}} \right)}^3} - 3\left( {\frac{y}{x}} \right)} \right]}}$ as it is $\phi \left( {\frac{y}{x}} \right)$ it is homogenous
Let $\frac{y}{x} = v$
y = v x
Differentiate w.r.t. 'x'
$\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$ Replacing value we get
$v + x\frac{{dv}}{{dx}} = \frac{{1 - 3{v^2}}}{{{v^3} - 3v}}$ $x\frac{{dv}}{{dx}} = \frac{{1 - 3{v^2}}}{{{v^3} - 3v}} - v$ $x\frac{{dv}}{{dx}} = \frac{{1 - \cancel{3{v^2}} - {v^4} + \cancel{ 3{v^2}}}}{{{v^3} - 3v}}$ $x\frac{{dv}}{{dx}} = \frac{{1 - {v^4}}}{{{v^3} - 3v}}$ $\frac{{{v^3} - 3v}}{{{v^4} - 1}}dv = - \frac{{dx}}{x}$ $\int {\frac{{{v^3}}}{{{v^4} - 1}}dv} - 3\int {\frac{v}{{{v^4} - 1}}dv = - \log \left( x \right) + \log c}$ Let V4 - 1 = t
4v3 dv = dt
${v^3}dv = \frac{{dt}}{4}$ Let v2 = u
2vdv = du
$vdv = \frac{{du}}{2}$ $\int {\frac{{\frac{{dt}}{4}}}{t} - 3\int {\frac{{\frac{{du}}{2}}}{{{u^2} - 1}} = \log \left( {\frac{c}{x}} \right)} }$ $\frac{1}{4}\log \left| t \right| - \frac{3}{2} \cdot \frac{1}{2}\log \left| {\frac{{u - 1}}{{u + 1}}} \right| = \log \left( {\frac{c}{x}} \right)$ $\frac{1}{4}\log \left| {{v^4} - 1} \right| - \frac{3}{2}\log \left| {\frac{{{v^2} - 1}}{{{v^2} + 1}}} \right| = \log \left( {\frac{c}{x}} \right)$ $\log \left| {{v^4} - 1} \right| - 6\log \left| {\frac{{{v^2} - 1}}{{{v^2} + 1}}} \right| = \log \left( {\frac{{{c^4}}}{{{x^4}}}} \right)$ $\log \left| {\left( {{v^4} - 1} \right)\left[ {\frac{{{{\left( {{v^2} - 1} \right)}^6}}}{{{{\left( {{v^2} + 1} \right)}^6}}}} \right]} \right| = \log \left( {\frac{{{c^4}}}{{{x^4}}}} \right)$ c is constant thus c4 = c $\log \left| {\frac{{\left( {{v^4} - 1} \right)\left( \cancel{{{v^2} + 1}} \right){{\left( {{v^2} - 1} \right)}^3}}}{{{{\left( {{v^2} + 1} \right)}\cancel{^3}^2}}}} \right| = \log \frac{c}{{{x^4}}}$ $\log \frac{{{{\left( {{v^2} - 1} \right)}^4}}}{{{{\left( {{v^2} + 1} \right)}^2}}} = \log \frac{c}{{{x^4}}}$ $\frac{{{{\left( {{v^2} - 1} \right)}^4}}}{{{{\left( {{v^2} + 1} \right)}^2}}} = \frac{c}{{{x^4}}}$ ${\left( {\frac{{{y^2}}}{{{x^2}}} - 1} \right)^4}{x^4} = c{\left( {\frac{{{y^2}}}{{{x^2}}} + 1} \right)^2}$ ${\left( {\frac{{{y^2} - {x^2}}}{{{x^2}}}} \right)^4}{x^4} = c{\left( {\frac{{{y^2}}}{{{x^2}}} + 1} \right)^2}$ $\frac{{{{\left( {{y^2} - {x^2}} \right)}^4}}}{{{x\cancel{^8}^4}}}{\cancel{x^4}} = c{\left( {\frac{{{y^2} + {x^2}}}{{{x^2}}}} \right)^2}$ $\frac{{{{\left( {{y^2} - {x^2}} \right)}^4}}}{{{x^4}}} = \frac{{c{{\left( {{y^2} + {x^2}} \right)}^2}}}{{{x^4}}}$ √c = c1 $\left( {{y^2} - {x^2}} \right) = {c_1}\left( {{x^2} + {y^2}} \right)$

Question (5)

From the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Solution

Let radius of circle be 'r' as it touches both axis in 1st quadrant, coordinate of centre C(r,r)

Equation of circle is
(x-r)2 + (y-r)2 = r2
Differentiate w.r.t x
${\rm{2(x - r) + 2(y - r)}}\frac{{dy}}{{dx}} = 0$ ${\rm{(x - r) + (y - r)}}\frac{{dy}}{{dx}} = 0$ $x - r + y{y_1} - r{y_1} = 0$ $x + y{y_1} = r\left( {1 + {y_1}} \right)$ $r = \frac{{x + y{y_1}}}{{1 + {y_1}}}$ Replacing value of 'r'
${\left( {x - \frac{{x + y{y_1}}}{{1 + {y_1}}}} \right)^2} + {\left( {y - \frac{{x + y{y_1}}}{{1 + {y_1}}}} \right)^2} = {\left( {\frac{{x + y{y_1}}}{{1 + {y_1}}}} \right)^2}$ ${\left( {x{y_1} + x - x - y{y_1}} \right)^2} + {\left( {y{y_1} + y - x - y{y_1}} \right)^2} = {\left( {x + y{y_1}} \right)^2}$ ${\left( {x - y} \right)^2}y_1^2 + {\left( {x - y} \right)^2} = {\left( {x + y{y_1}} \right)^2}$ ${\left( {x - y} \right)^2}\left( {1 + y_1^2} \right) = {\left( {x + y{y_1}} \right)^2}$

Question (6)

Find the general solution of the differential equation $\frac{{dy}}{{dx}} + \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} = 0$

Solution

$\frac{{dy}}{{dx}} + \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} = 0$ $\frac{{dy}}{{dx}} = - \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}}$ $\frac{{dy}}{{\sqrt {1 - {y^2}} }} = - \frac{{dx}}{{\sqrt {1 - {x^2}} }}$ $\int {\frac{{dy}}{{\sqrt {1 - {y^2}} }}} = - \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}$ ${\sin ^{ - 1}}y = - {\sin ^{ - 1}}x + c$ ${\sin ^{ - 1}}y + {\sin ^{ - 1}}x = c$

Question (7)

Show that the general solution of the differential equation
$\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0$
is given by $\left( {x + y + 1} \right) = A\left( {1 - x - y - 2xy} \right)$, where A is parameter

Solution

$\frac{{dy}}{{dx}} + \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}} = 0$ $\frac{{dy}}{{dx}} = - \frac{{{y^2} + y + 1}}{{{x^2} + x + 1}}$ $\frac{{dy}}{{{y^2} + y + 1}} = - \frac{{dx}}{{{x^2} + x + 1}}$ $\int {\frac{{dy}}{{{y^2} + y + 1}}} = - \int {\frac{{dx}}{{{x^2} + x + 1}}}$ $\int {\frac{{dy}}{{{y^2} + y + \frac{1}{4} + \frac{3}{4}}}} = - \int {\frac{{dx}}{{{x^2} + x + \frac{1}{4} + \frac{3}{4}}}}$ $\int {\frac{{dy}}{{{{\left( {y + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} = - \int {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}}$ $\frac{1}{{\frac{{\sqrt 3 }}{2}}}{\tan ^{ - 1}}\left( {\frac{{y + \frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}}} \right) = - \frac{1}{{\frac{{\sqrt 3 }}{2}}}{\tan ^{ - 1}}\left( {\frac{{x + \frac{1}{2}}}{{\sqrt {\frac{3}{2}} }}} \right) + \frac{{2c}}{{\sqrt 3 }}$ ${\tan ^{ - 1}}\left( {\frac{{2y + 1}}{{\sqrt 3 }}} \right) = - {\tan ^{ - 1}}\left( {\frac{{2x + 1}}{{\sqrt 3 }}} \right) + \frac{{2c}}{{\sqrt 3 }}$ ${\tan ^{ - 1}}\left( {\frac{{2y + 1}}{{\sqrt 3 }}} \right) + {\tan ^{ - 1}}\left( {\frac{{2x + 1}}{{\sqrt 3 }}} \right) = \frac{{2c}}{{\sqrt 3 }}$ ${\tan ^{ - 1}}\left( {\frac{{\frac{{2y + 1}}{{\sqrt 3 }} + \frac{{2x + 1}}{{\sqrt 3 }}}}{{1 - \frac{{\left( {2y + 1} \right)\left( {2x + 1} \right)}}{{\sqrt 3 \cdot \sqrt 3 }}}}} \right) = \frac{{2c}}{{\sqrt 3 }}$ $\frac{{2x + 2y + 2}}{{\sqrt 3 }} \times \frac{3}{{3 - 4xy - 2y - 2x - 1}} = \tan \left( {\frac{{2c}}{{\sqrt 3 }}} \right)$ $\frac{{\cancel{2}\left( {x + y + 1} \right)\sqrt 3 }}{{\cancel{2}\left( {1 - x - y - xy} \right)}} = \tan \left( {\frac{{2c}}{{\sqrt 3 }}} \right)$ $\therefore x + y + 1 = \frac{1}{{\sqrt 3 }}\tan \left( {\frac{{2c}}{{\sqrt 3 }}} \right)\left( {1 - x - y - xy} \right)$ $text{as}\quad \frac{1}{{\sqrt 3 }}\tan \left( {\frac{{2c}}{{\sqrt 3 }}} \right) = constant = A$ $x + y + 1 = A\left( {1 - x - y - xy} \right)$

Question (8)

Find the equation of the curve passing through the point $\left( {0,\frac{\pi }{4}} \right)$ whose diffrential equation is sinxcosydx + cosx siny dy = 0

Solution

Curve passing through $\left( {0,\frac{\pi }{4}} \right)$

${\rm{sin}}\left( {\rm{x}} \right){\rm{cos}}\left( {\rm{y}} \right){\rm{dx + cos}}\left( {\rm{x}} \right){\rm{sin}}\left( {\rm{y}} \right){\rm{dy = 0}}$ ${\rm{cos}}\left( {\rm{x}} \right){\rm{sin}}\left( {\rm{y}} \right){\rm{dy = - sin}}\left( {\rm{x}} \right){\rm{cos}}\left( {\rm{y}} \right){\rm{dx }}$ $\frac{{\sin ydy}}{{\cos y}} = \frac{{ - \sin xdx}}{{\cos x}}$ $\tan ydy = - \tan xdx$ $\int {\tan ydy} = - \int {\tan xdx}$ $\log \left| {\sec y} \right| = - \log \left| {\sec x} \right| + logc$ $\log \left| {\sec y} \right| = \log \left| {\frac{c}{{\sec x}}} \right|$ $\Rightarrow \sec y = \frac{c}{{\sec x}}$ $\sec x\sec y = c$ $\therefore \cos x\cos y = \frac{1}{c} = k$ it passes through $\left( {0,\frac{\pi }{4}} \right)$
$\therefore \cos 0\cos \frac{\pi }{4} = \frac{1}{c}$ $1 \cdot \frac{1}{{\sqrt 2 }} = \frac{1}{c}$ $c = \sqrt 2$ $\therefore \cos x\cos y = \sqrt 2$

Question (9)

Find the particular solution of the differential equation
$\left( {1 + {e^{2x}}} \right)dy + \left( {1 + {y^2}} \right){e^x}dx = 0$ , given that y=1 when x=0

Solution

$\left( {1 + {e^{2x}}} \right)dy = - \left( {1 + {y^2}} \right){e^x}dx$ $\frac{{dy}}{{1 + {y^2}}} = - \frac{{{e^x}}}{{1 + {{\left( {{e^x}} \right)}^2}}}dx$ $\int {\frac{{dy}}{{1 + {y^2}}}} = - \int {\frac{{{e^x}}}{{1 + {{\left( {{e^x}} \right)}^2}}}dx}$ ${\tan ^{ - 1}}y = - {\tan ^{ - 1}}{e^x} + c$ ${\tan ^{ - 1}}y + {\tan ^{ - 1}}{e^x} = c$ $y = 1, \quad \text{and} x = 0$ ${\tan ^{ - 1}}1 + {\tan ^{ - 1}}{e^0} = c$ $c = \frac{\pi }{2}$ $\therefore {\tan ^{ - 1}}y + {\tan ^{ - 1}}{e^x} = \frac{\pi }{2}$

Question (10)

Solve the differential equation $y{e^{\frac{x}{y}}}dx = \left( {x{e^{\frac{x}{y}}} + {y^2}} \right)dy\left( {y \ne 0} \right)$

Solution

$y{e^{\frac{x}{y}}}dx = \left( {x{e^{\frac{x}{y}}} + {y^2}} \right)dy$ ${e^{\frac{x}{y}}}dx = \left( {\frac{x}{y}{e^{\frac{x}{y}}} + y} \right)dy$ $\frac{{dx}}{{dy}} = \frac{x}{y} + \frac{y}{{{e^{\frac{x}{y}}}}}$ $\text{Let} \quad\frac{x}{y} = v$ $x = vy$ Differentiate w.r.t y $\frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}$ $\cancel{v} + y\frac{{dv}}{{dy}} = \cancel{v} + \frac{y}{{{e^v}}}$ $\cancel{y}\frac{{dv}}{{dy}} = \frac{\cancel{y}}{{{e^v}}}$ ${e^v}dv = dy$ $\int {{e^v}dv} = \int {dy}$ ${e^v} = y + c$ ${e^{\frac{x}{y}}} = y + c$

Question (11)

Find a particular solution of the differential equation (x - y) (dx + dy) = dx - dy.
given that y = -1, when x = 0 ( Hint put x - y = t )

Solution

$\left( {x - y} \right)\left( {dx + dy} \right) = dx - dy$ $\left( {x - y + } \right)dy = \left( {1 - x + y} \right)dx$ $\frac{{dy}}{{dx}} = \frac{{1 - \left( {x - y} \right)}}{{x - y + 1}}$ Let x - y = t
$1 - \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$
$1 - \frac{{1 - t}}{{t + 1}} = \frac{{dt}}{{dx}}$ $\frac{{t + \cancel{1} - \cancel{1} + t}}{{t + 1}} = \frac{{dt}}{{dx}}$ $\frac{{2t}}{{t + 1}} = \frac{{dt}}{{dx}}$ $\frac{{t + 1}}{t}dt = 2dx$ $\int {dt} + \int {\frac{1}{t}dt = 2\int {dx} }$ $t + \log t = 2x + c$ $x - y + \log \left| {x - y} \right| = 2x + c$ $\log \left| {x - y} \right| = x + y + c$ when x = 0 and y = -1
$\log 1 = 0 - 1 + c$ $\log \left| {x - y} \right| = x + y + 1$

Question (12)

Solve the differential equation
$\left[ {\frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}} \right]\frac{{dy}}{{dx}} = 1\left( {x \ne 0} \right)$

Solution

$\left[ {\frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}} \right]\frac{{dx}}{{dy}} = 1$ $\frac{{dy}}{{dx}} = \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }} - \frac{y}{{\sqrt x }}$ $\frac{{dy}}{{dx}} + \frac{y}{{\sqrt x }} = \frac{{{e^{ - 2\sqrt x }}}}{{\sqrt x }}$ $p = \frac{1}{{\sqrt x }}$ $I.F = {e^{\int {pdx} }}$ $I.F = {e^{\int {\frac{1}{{\sqrt x }}dx} }}$ $I.F = {e^{\int {{x^{\frac{{ - 1}}{2}}}dx} }}$ $I.F = {e^{2{x^{\frac{1}{2}}}}}$ $I.F = {e^{2\sqrt x }}$ ${e^{2\sqrt x }}\frac{{dy}}{{dx}} + {e^{2\sqrt x }}\frac{y}{{\sqrt x }} = \frac{{{e^{ - 2\sqrt x }} \times {e^{2\sqrt x }}}}{{\sqrt x }}$ $\frac{d}{{dx}}\left( {y{e^{2\sqrt x }}} \right) = \frac{1}{{\sqrt x }}$ $\int {\frac{d}{{dx}}\left( {y{e^{2\sqrt x }}} \right)dx} = \int {\frac{1}{{\sqrt x }}dx}$ $y{e^{2\sqrt x }} = 2\sqrt x + c$

Question (13)

Find the paraticular solution of the differential equation $\frac{{dy}}{{dx}} + y\cot x4x cosec(x)\left( {x \ne 0} \right)$ given that y=0, when $x = \frac{\pi }{2}$

Solution

$\frac{{dy}}{{dx}} + y\cot x = 4xcosec(x)$ comparing we get
p = cotx $I.F. = {e^{\int {pdx} }}$ $I.F. = {e^{\int {\cot xdx} }}$ $I.F. = {e^{\log \left| {\sin x} \right|}}$ $I.F. = \sin {x^{\log _e^e}}$ $I.F. = \sin x$ Mltiply by sinx $\sin x\frac{{dy}}{{dx}} + y\cot x\sin x = 4x\sin x cosec(x)$ $\frac{d}{{dx}}\left( {y\sin x} \right) = 4x$ $\int {\frac{d}{{dx}}\left( {y\sin x} \right)dx} = 4\int {xdx}$ $y\sin x = 4\frac{{{x^2}}}{2} + c$ $y\sin x = 2{x^2} + c$ If y = 0, $x = \frac{\pi }{2}$
$0 = \cancel{2}\left( {\frac{{{\pi ^2}}}{\cancel{4}}} \right) + c$ $c = \frac{{{\pi ^2}}}{2}$ $y\sin x = 2{x^2} + \frac{{{\pi ^2}}}{2}$

Question (14)

Find a particular solution of the differential equation $\left( {x + 1} \right)\frac{{dy}}{{dx}} = 2{e^{ - y}} - 1$ given that y = 0 when x = 0

Solution

$\left( {x + 1} \right)\frac{{dy}}{{dx}} = 2{e^{ - y}} - 1$ y = 0 and y = 0 $\frac{{dy}}{{2{e^{ - y}} - 1}} = \frac{{dx}}{{x + 1}}$ $\frac{{dy}}{{\frac{2}{{{e^y}}} - 1}} = \frac{{dx}}{{x + 1}}$ $\frac{{{e^y}dy}}{{2 - {e^y}}} = \frac{{dx}}{{x + 1}}$ $\int {\frac{{{e^y}dy}}{{2 - {e^y}}}} = \int {\frac{{dx}}{{x + 1}}}$ $2 - {e^y} = t$ $- {e^y}dy = dt$ $\Rightarrow {e^y}dy = - dt$ $\int { - \frac{{dt}}{t} = \int {\frac{{dx}}{{x + 1}}} }$ $- \log t = \log \left| {x + 1} \right| + \log c$ $\log \left| {\frac{1}{t}} \right| = \log \left| {c\left( {x + 1} \right)} \right|$ $c\left( {x + 1} \right) = \frac{1}{t}$ $ct\left( {x + 1} \right) = 1$ $c\left( {2 - {e^y}} \right)\left( {x + 1} \right) = 1$ y= 0 and x =0
$c\left( {2 - 1} \right)\left( 1 \right) = 1$ $c = 1$ Replacing value of c $\left( {2 - {e^y}} \right)\left( {x + 1} \right) = 1$

Question (15)

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?

Solution

Population is 'x'
$\frac{{dx}}{{dt}} \propto x$ $\frac{{dx}}{{dt}} = kx\left( {k \ne 0} \right)$ $\thereffore \frac{{dx}}{x} = kdt$ $\int {\frac{{dx}}{x}} = k\int {dt}$ $\log \left| x \right| = kt + \log c$ $\log \left| {\frac{x}{c}} \right| = kt$ $\frac{x}{c} = {e^{kt}}$ $x = c{e^{kt}}$ When x = 20,000 , t=0
$20000 = c{e^{k\left( 0 \right)}} = c$ $x = 20000{e^{kt}}$ x=25,000 and t = 5
$25000 = 20000{e^{k5}}$ ${e^{k5}} = \frac{5}{4}$ When t = 10 and x = ? $x = 20000{e^{kt}}$ $x = 20000{e^{10k}}$ $x = 20000{\left( {{e^{5k}}} \right)^2}$ $x = 20000{\left( {\frac{5}{4}} \right)^2}$ $x = \cancel{20000}^1250\left( {\frac{{25}}{{\cancel{16}}}} \right)$ $x = 31250$

Question (16)

The general solution of the differential equation $\frac{{ydx - xdy}}{y} = 0$ is
(A) xy = C     (B) x = Cy2
(C) y = Cx     (D) y = Cx2

Solution

$\frac{{ydx - xdy}}{y} = 0$ $ydx = xdy$ $\frac{{dy}}{y} = \frac{{dx}}{x}$ $\int {\frac{{dy}}{y}} = \int {\frac{{dx}}{x}}$ $\log y = \log x + \log c$ $\log y = \log \left( {cx} \right)$ $y = ca$ Option (C) is correct

Question (17)

The genral solution of differtial equation of the type $\frac{{dx}}{{dy}} + {P_1}x = {Q_1}$ is
$\left( A \right)y{e^{\int {{P_1}dy} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dy} }}} \right)} dy + C$
$\left( B \right)y{e^{\int {{P_1}dx} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dx} }}} \right)} dx + C$
$\left( C \right)x{e^{\int {{P_1}dy} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dy} }}} \right)} dy + C$
$\left( D \right)x{e^{\int {{P_1}dx} }} = \int {\left( {{Q_1}{e^{\int {{P_1}dx} }}} \right)} dx + C$

Solution

$\frac{{dx}}{{dy}} + {P_1}x = {Q_1}$ $M.F = {e^{\int {{P_1}dy} }}$ $y{e^{\int {{P_1}dy} }} = \int {{Q_1}{e^{\int {{P_1}dy} }}dy + c}$ Option (A) is correct

Question (18)

The general solution of the differential equation
${e^x}dy + \left( {y{e^x} + 2x} \right)dx = 0$ is
$\left( A \right)x{e^y} + {x^2} = C$
$\left( B \right)x{e^y} + {y^2} = C$
$\left( C \right)y{e^x} + {x^2} = C$
$\left( D \right)y{e^y} + {x^2} = C$

Solution

${e^x}dy + \left( {y{e^x} + 2x} \right)dx = 0$ ${e^x}dy = - \left( {y{e^x} + 2x} \right)dx$ $\frac{{dy}}{{dx}} = - \frac{{y{e^x} + 2x}}{{{e^x}}}$ $\frac{{dy}}{{dx}} = - y + 2x{e^{ - x}}$ $\frac{{dy}}{{dx}} + y = 2x{e^{ - x}}$ P = 1 $MF = {e^{\int {Pdx} }}$ $MF = {e^x}$ ${e^x}\frac{{dy}}{{dx}} + y{e^x} = 2x{e^{ - x}}{e^x}$ $\int {\frac{d}{{dx}}\left( {y{e^x}} \right)dx} = 2\int {xdx}$ $y{e^x} = \cancel{2}\frac{{{x^2}}}{\cancel{2}} + c$ $y{e^x} = {x^2} + c$ Option (C) iis correct