Question (1)
For each of the differential 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.$(i)xy = a{e^x} + b{e^{ - x}} + {x^2}\quad :x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2 = 0$
$(ii)y = {e^x}\left( {a\cos x + b\sin x} \right)\quad :\frac{{{d^2}y}}{{d{x^2}}} - 2\frac{{dy}}{{dx}} + 2y = 0$
$(iii)y = x\sin 3x \quad :\frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x = 0$ $(iv){x^2} = 2{y^2}\log y\quad :\left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy = 0$
Solution
$(i)xy = a{e^x} + b{e^{ - x}} + {x^2}\quad :x\frac{{{d^2}y}}{{d{x^2}}} + 2\frac{{dy}}{{dx}} - xy + {x^2} - 2 = 0$xy = a{e^x} + b{e^{ - x}} + {x^2}
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'
\[\frac{{{d^2}y}}{{d{x^2}}} = 3\left[ {x\left( { - \sin 3x} \right)3 + \cos 3x\left( 1 \right)} \right] + \cos 3x\left( 3 \right)\] \[\frac{{{d^2}y}}{{d{x^2}}} = 3\left( { - 3x\sin 3x + \cos 3x} \right) + 3\cos 3x\] \[\frac{{{d^2}y}}{{d{x^2}}} = - 9x\sin 3x + 3\cos 3x + 3\cos 3x\] \[\frac{{{d^2}y}}{{d{x^2}}} + 9y - 6\cos 3x = 0\]
$(iv){x^2} = 2{y^2}\log y\quad :\left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} - xy = 0$
\[{x^2} = 2{y^2}\log y\] 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 homogenousLet $\frac{y}{x} = v$
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
\[vdv = \frac{{du}}{2}\] \[\int {\frac{{\frac{{dt}}{4}}}{t} - 3\int {\frac{{\frac{{du}}{2}}}{{{u^2} - 1}} = \log \left( {\frac{c}{x}} \right)} } \] \[\frac{1}{4}\log \left| t \right| - \frac{3}{2} \cdot \frac{1}{2}\log \left| {\frac{{u - 1}}{{u + 1}}} \right| = \log \left( {\frac{c}{x}} \right)\] \[\frac{1}{4}\log \left| {{v^4} - 1} \right| - \frac{3}{4}\log \left| {\frac{{{v^2} - 1}}{{{v^2} + 1}}} \right| = \log \left( {\frac{c}{x}} \right)\] \[\log \left| {{v^4} - 1} \right| - 3\log \left| {\frac{{{v^2} - 1}}{{{v^2} + 1}}} \right| = \log \left( {\frac{{{c^4}}}{{{x^4}}}} \right)\] \[\log \left| {\left( {{v^4} - 1} \right)\left[ {\frac{{{{\left( {{v^2} + 1} \right)}^3}}}{{{{\left( {{v^2} - 1} \right)}^3}}}} \right]} \right| = \log \left( {\frac{{{c^4}}}{{{x^4}}}} \right)\] \[\log \left| {\frac{{\left( {{v^2} - 1} \right)\left( {{v^2} + 1} \right){{\left( {{v^2} + 1} \right)}^3}}}{{{{\left( {{v^2} - 1} \right)}^3}}}} \right| = \log \frac{c}{{{x^4}}}\] \[\log \frac{{{{\left( {{v^2} + 1} \right)}^4}}}{{{{\left( {{v^2} - 1} \right)}^2}}} = \log \frac{c^4}{{{x^4}}}\] \[\frac{{{{\left( {{v^2} + 1} \right)}^4}}}{{{{\left( {{v^2} - 1} \right)}^2}}} = \frac{c^4}{{{x^4}}}\] \[{\left( {\frac{{{y^2}}}{{{x^2}}} + 1} \right)^4}{x^4} = c^4{\left( {\frac{{{y^2}}}{{{x^2}}} - 1} \right)^2}\] \[{\left( {\frac{{{y^2} + {x^2}}}{{{x^2}}}} \right)^4}{x^4} = c^4{\left( {\frac{{{y^2}}}{{{x^2}}} - 1} \right)^2}\] \[\frac{{{{\left( {{y^2} + {x^2}} \right)}^4}}}{{{x\cancel{^8}^4}}}{\cancel{x^4}} = c^4{\left( {\frac{{{y^2} - {x^2}}}{{{x^2}}}} \right)^2}\] \[\frac{{{{\left( {{y^2} + {x^2}} \right)}^4}}}{{{x^4}}} = \frac{{c^4{{\left( {{x^2} - {y^2}} \right)}^2}}}{{{x^4}}}\] Taking square root on both side
\[{\left( {{x^2} + {y^2}} \right)^2} = {c^2}\left( {{x^2} - {y^2}} \right)\] \[\left( {{x^2} - {y^2}} \right) = \frac{1}{{{c^2}}}{\left( {{x^2} + {y^2}} \right)^2}\] let 1/c^2 = C
\[\left( {{x^2} - {y^2}} \right) = C{\left( {{x^2} + {y^2}} \right)^2}\]
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 = 0Solution
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
\[\left( {x - y} \right)\left( {dx + dy} \right) = dx - dy\] \[\left( {x - y + 1} \right)dy = \left( {1 - x + y} \right)dx\] \[\frac{{dy}}{{dx}} = \frac{{1 - \left( {x - y} \right)}}{{x - y + 1}}\] Let x - y = t$1 - \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$
Replacing the value we get
\[1 - \frac{{dt}}{{dx}} = \frac{{1 - t}}{{1 + t}}\] \[1 - \frac{{1 - t}}{{t + 1}} = \frac{{dt}}{{dx}}\] \[\frac{{t + \cancel{1} - \cancel{1} + t}}{{t + 1}} = \frac{{dt}}{{dx}}\] \[\frac{{2t}}{{t + 1}} = \frac{{dt}}{{dx}}\] \[\frac{{t + 1}}{t}dt = 2dx\] \[\int {dt} + \int {\frac{1}{t}dt = 2\int {dx} } \] \[t + \log t = 2x + c\] \[x - y + \log \left| {x - y} \right| = 2x + c\] \[\log \left| {x - y} \right| = x + y + c\] 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 \[\frac{{dy}}{{dx}} + y\cot x = 4x cosec(x)\left( {x \ne 0} \right)\] given that y=0, when $x = \frac{\pi }{2}$Solution
\[\frac{{dy}}{{dx}} + y\cot x = 4xcosec(x)\] comparing we getp = 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 \[\sin x\frac{{dy}}{{dx}} + y\cot x\sin x = 4x\sin x cosec(x)\] \[\sin x\frac{{dy}}{{dx}} + y\frac{{\cos x}}{{\sin x}}x\sin x = 4x\sin xcosec(x)\] \[\sin x\frac{{dy}}{{dx}} + y\cos x = 4x\sin xcosec(x)\] \[\frac{d}{{dx}}\left( {y\sin x} \right) = 4x\] \[\int {\frac{d}{{dx}}\left( {y\sin x} \right)dx} = 4\int {xdx} \] \[y\sin x = 4\frac{{{x^2}}}{2} + c\] \[y\sin x = 2{x^2} + c\] 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 = 0Solution
\[\left( {x + 1} \right)\frac{{dy}}{{dx}} = 2{e^{ - y}} - 1\] y = 0 and x = 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'\[\frac{{dx}}{{dt}} \propto x\] \[\frac{{dx}}{{dt}} = kx\left( {k \ne 0} \right)\] \[\therefore \frac{{dx}}{x} = kdt\] \[\int {\frac{{dx}}{x}} = k\int {dt} \] \[\log \left| x \right| = kt + \log c\] \[\log \left| {\frac{x}{c}} \right| = kt\] \[\frac{x}{c} = {e^{kt}}\] \[x = c{e^{kt}}\] 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
\[\frac{{ydx - xdy}}{y} = 0\] \[ydx = xdy\] \[\frac{{dy}}{y} = \frac{{dx}}{x}\] \[\int {\frac{{dy}}{y}} = \int {\frac{{dx}}{x}} \] \[\log y = \log x + \log c\] \[\log y = \log \left( {cx} \right)\] \[y = cx\] Option (C) is correctQuestion (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 correctQuestion (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