12th NCERT Differential Equation Exercise 9.5 Number of questions 17
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In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them

Question (1)

(x2 +xy) dy = (x2+y2) dx

Solution

\[\frac{{dy}}{{dx}} = \frac{{{x^2} + {y^2}}}{{{x^2} + xy}}\] \[\frac{{dy}}{{dx}} = \frac{{{x^2}\left[ {1 + {{\left( {\frac{y}{x}} \right)}^2}} \right]}}{{{x^2}\left[ {1 + \left( {\frac{y}{x}} \right)} \right]}}\] \[\frac{{dy}}{{dx}} = \frac{{1 + {{\left( {\frac{y}{x}} \right)}^2}}}{{1 + \left( {\frac{y}{x}} \right)}}\quad - - - \left( 1 \right)\] let y/x = v
∴ y = vx
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] Replacing the value of dy/dx and y/x = v in equation (1) \[v + x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{1 + v}}\] \[x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{1 + v}} - v\] \[x\frac{{dv}}{{dx}} = \frac{{1 + \require{cancel}\cancel{{v^2}} - v - \cancel{{v^2}}}}{{1 + v}}\] \[x\frac{{dv}}{{dx}} = \frac{{1 - v}}{{v + 1}}\] \[x\frac{{dv}}{{dx}} = - \frac{{v - 1}}{{v + 1}}\] \[\frac{{v + 1}}{{v - 1}}dv = - \frac{{dx}}{x}\] \[\int {\frac{{v + 1}}{{v - 1}}dv} = - \int {\frac{{dx}}{x}} \] \[\int {\frac{{v - 1 + 2}}{{v - 1}}dv} = - \int {\frac{{dx}}{x}} \] \[\int {dv} + 2\int {\frac{1}{{v - 1}}dv = - \log x + \log c} \] \[v + 2\log \left| {v - 1} \right| = \log \frac{c}{x}\] \[\frac{y}{x} + 2\log \left| {\frac{y}{x} - 1} \right| = \log \frac{c}{x}\] \[\frac{y}{x} + 2\log \left( {y - x} \right) - 2\log x = \log \frac{c}{x}\] \[\frac{y}{x} + \log {\left( {y - x} \right)^2} = \log cx\] \[\frac{y}{x} = \log \left| {\frac{{cx}}{{{{\left( {y - x} \right)}^2}}}} \right|\] \[\frac{{cx}}{{{{\left( {y - x} \right)}^2}}} = {e^{\frac{y}{x}}}\] \[{\left( {y - x} \right)^2} = \frac{{cx}}{{{e^{\frac{y}{x}}}}}\] \[{\left( {y - x} \right)^2} = cx{e^{\frac{{ - y}}{x}}}\]

Question (2)

\[y' = \frac{{x + y}}{x}\]

Solution

\[\frac{{dy}}{{dx}} = \frac{x}{x} + \frac{y}{x}\] \[\frac{{dy}}{{dx}} = 1 + \frac{y}{x}\] let y/x= v
y = vx
differentiate with respect to x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] Replacing the values we get
\[\cancel{v} + x\frac{{dv}}{{dx}} = 1 + \cancel{v}\] \[dv = \frac{{dx}}{x}\] \[\int {dv} = \int {\frac{{dx}}{x}} \] v=logx+c
y/x = log x + c
y = xlogx + cx

Question (3)

(x-y)dy - (x+y)dx=0

Solution

\[\frac{{dy}}{{dx}} = \frac{{x + y}}{{x - y}}\] \[\frac{{dy}}{{dx}} = \frac{{x\left( {1 + \frac{y}{x}} \right)}}{{x\left( {1 - \frac{y}{x}} \right)}}\] Let y/x = v
y = v x
differentiate with respect x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] Replacing the values we get
\[v + x\frac{{dv}}{{dx}} = \frac{{1 + v}}{{1 - v}}\] \[x\frac{{dv}}{{dx}} = \frac{{1 + v}}{{1 - v}} - v\] \[x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{1 - v}}\] \[x\frac{{dv}}{{dx}} = \frac{{1 + {v^2}}}{{ - \left( {v - 1} \right)}}\] \[\frac{{\left( {v - 1} \right)dv}}{{{v^2} + 1}} = - \int {\frac{{dx}}{x}} \] \[\int {\frac{v}{{{v^2} + 1}}dv - \int {\frac{1}{{{v^2} + 1}}dv} = - \log x + \log c} \] \[\frac{1}{2}\int {\frac{{2v}}{{{v^2} + 1}}dv - {{\tan }^{ - 1}}v = \log \left( {\frac{c}{x}} \right)} \] \[\frac{1}{2}\log \left| {{v^2} + 1} \right| - {\tan ^{ - 1}}v = \log c - \log x\] \[\frac{1}{2}\log \left| {\frac{{{y^2}}}{{{x^2}}} + 1} \right| - {\tan ^{ - 1}}\frac{y}{x} = \log c - \log x\] \[\frac{1}{2}\log \left( {{y^2} + {x^2}} \right) - \frac{1}{2}\log {x^2} - {\tan ^{ - 1}}\frac{y}{x} = \log c - \log x\] \[\frac{1}{2}\log \left( {{y^2} + {x^2}} \right) - \cancel{\log x} - {\tan ^{ - 1}}\frac{y}{x} = \log c - \cancel{\log x}\] \[\frac{1}{2}\log \left( {{y^2} + {x^2}} \right) - {\tan ^{ - 1}}\frac{y}{x} = c\]

Question (4)

(x2 -y2)dx + 2xy dy = 0

Solution

2xydy = (y2 - x2) dx \[\frac{{dy}}{{dx}} = \frac{{{y^2} - {x^2}}}{{2xy}}\] \[\frac{{dy}}{{dx}} = \frac{{{\cancel{x^2}}\left( {\frac{{{y^2}}}{{{x^2}}} - 1} \right)}}{{2{\cancel{x^2}}\frac{y}{x}}}\] let y/x = v
y = vx
differentiate with respect to x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] Replacing the values \[v + x\frac{{dv}}{{dx}} = \frac{{{v^2} - 1}}{{2v}}\] \[x\frac{{dv}}{{dx}} = \frac{{{v^2} - 1}}{{2v}} - v\] \[x\frac{{dv}}{{dx}} = \frac{{{v^2} - 1 - 2{v^2}}}{{2v}}\] \[x\frac{{dv}}{{dx}} = - \frac{{\left( {{v^2} + 1} \right)}}{{2v}}\] \[\frac{{2v}}{{{v^2} + 1}}dv = - \frac{{dx}}{x}\] \[\int {\frac{{2v}}{{{v^2} + 1}}dv} = - \int {\frac{{dx}}{x}} \] log|v2+1| = -logx + log c
log|v2+1| = log c/x
v2+1 = c/x
\[x\left( {\frac{{{y^2}}}{{{x^2}}} + 1} \right) = c\] \[\cancel{x}\left( {\frac{{{y^2} + {x^2}}}{{{x^\cancel{2}}}}} \right) = c\] x2 + y2 = cx

Question (5)

\[{x^2}\frac{{dy}}{{dx}} = {x^2} - 2{y^2} + xy\]

Solution

\[\frac{{dy}}{{dx}} = \frac{{{x^2} - 2{y^2} + xy}}{{{x^2}}}\] \[\frac{{dy}}{{dx}} = 1 - 2{\left( {\frac{y}{x}} \right)^2} + \left( {\frac{y}{x}} \right)\] let y/x = v
y = vx
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] \[\cancel{v} + x\frac{{dv}}{{dx}} = 1 - 2{v^2} + \cancel{v}\] \[x\frac{{dv}}{{dx}} = 1 - 2{v^2}\] \[x\frac{{dv}}{{dx}} = - \left( {2{v^2} - 1} \right)\] \[\int {\frac{{dv}}{{2{v^2} - 1}} = - \int {\frac{{dx}}{x}} } \] \[\frac{1}{2}\int {\frac{{dv}}{{{v^2} - \frac{1}{2}}} = - \int {\frac{{dx}}{x}} } \] \[\frac{1}{{2 \cdot \frac{2}{{\sqrt 2 }}}}\log \left| {\frac{{v - \frac{1}{{\sqrt 2 }}}}{{v + \frac{1}{{\sqrt 2 }}}}} \right| = - \log x + \log c\] \[\frac{1}{{2 \cdot \frac{2}{{\sqrt 2 }}}}\log \left| {\frac{{\sqrt 2 v - 1}}{{\sqrt 2 v + 1}}} \right| = \log c - \log x\] \[\frac{1}{{2\sqrt 2 }}\log \left| {\frac{{\sqrt 2 y - x}}{{\sqrt 2 y + x}}} \right| = \log c - \log x\]

Question (6)

\[xdy - ydx = \sqrt {{x^2} + {y^2}} dx\]

Solution

\[xdy = \sqrt {{x^2} + {y^2}} dx + ydx\] \[xdy = \left( {\sqrt {{x^2} + {y^2}} + y} \right)dx\] \[\frac{{dy}}{{dx}} = \frac{{\sqrt {{x^2} + {y^2}} + y}}{x}dx\] \[\frac{{dy}}{{dx}} = \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} + \frac{y}{x}\] y/x = v
y = vx
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] Replacing the values we get
\[\cancel{v} + x\frac{{dv}}{{dx}} = \sqrt {1 + {v^2}} + \cancel{v}\] \[x\frac{{dv}}{{dx}} = \sqrt {1 + {v^2}} \] \[\frac{{dv}}{{\sqrt {1 + {v^2}} }} = \frac{{dx}}{x}\] \[\int {\frac{{dv}}{{\sqrt {1 + {v^2}} }}} = \int {\frac{{dx}}{x}} \] \[\log \left| {v + \sqrt {1 + {v^2}} } \right| = \log x + \log c\] \[\log \left| {v + \sqrt {1 + {v^2}} } \right| = \log cx\] \[v + \sqrt {1 + {v^2}} = cx\] \[\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} = cx\] \[\frac{y}{x} + \frac{{\sqrt {{x^2} + {y^2}} }}{x} = cx\] \[y + \sqrt {{x^2} + {y^2}} = c{x^2}\]

Question (7)

\[\left\{ {x\cos \left( {\frac{y}{x}} \right) + y\sin \left( {\frac{y}{x}} \right)} \right\}ydx = \left\{ {y\sin \left( {\frac{y}{x}} \right) - x\cos \left( {\frac{y}{x}} \right)} \right\}xdy\]

Solution

\[\left\{ {\cos \left( {\frac{y}{x}} \right) + \frac{y}{x}\sin \left( {\frac{y}{x}} \right)} \right\}\frac{y}{x}dx = \left\{ {\frac{y}{x}\sin \left( {\frac{y}{x}} \right) - \cos \left( {\frac{y}{x}} \right)} \right\}dy\] \[{\frac{{dy}}{{dx}} = \frac{{\left\{ {\cos \left( {\frac{y}{x}} \right) + \frac{y}{x}\sin \left( {\frac{y}{x}} \right)} \right\}\frac{y}{x}}}{{\frac{y}{x}\sin \left( {\frac{y}{x}} \right) - \cos \left( {\frac{y}{x}} \right)}}}\] Let y/x =v
y = vx
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] Replacing the values we get
\[x\frac{{dv}}{{dx}} = \frac{{\left( {\cos v + v\sin v} \right)v}}{{v\sin v - \cos v}} - v\] \[x\frac{{dv}}{{dx}} = \frac{{v\cos v + \cancel{{v^2}\sin v} - \cancel{{v^2}\sin v} + v\cos v}}{{v\sin v - \cos v}}\] \[x\frac{{dv}}{{dx}} = \frac{{2v\cos v}}{{v\sin v - \cos v}}\] \[\frac{{v\sin v - \cos v}}{{v\cos v}}dv = 2\frac{{dx}}{x}\] \[\int {\tan vdv - \int {\frac{1}{v}dv} = 2\int {\frac{{dx}}{x}} } \] log|sec v| - log V = 2log x + log c
\[\log \left| {\frac{{\sec v}}{v}} \right| = \log c{x^2}\] \[\frac{{\sec v}}{v} = c{x^2}\] \[\frac{\cancel{x}}{{\cos \left( {\frac{y}{x}} \right)y}} = c{x^\cancel{2}}\] \[xy\cos \left( {\frac{y}{x}} \right) = \frac{1}{c} = c\]

Question (8)

\[x\frac{{dy}}{{dx}} - y + x\sin \left( {\frac{y}{x}} \right) = 0\]

Solution

\[\frac{{dy}}{{dx}} - \frac{y}{x} + \sin \left( {\frac{y}{x}} \right) = 0\] Let y/x = v
y = vx
differentiate w.r.t x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] \[\cancel{v} + x\frac{{dv}}{{dx}} - \cancel{v} + \sin v = 0\] \[x\frac{{dv}}{{dx}} = - \sin v\] \[\frac{{dv}}{{\sin v}} = - \frac{{dx}}{x}\] \[\int {cosec vdv = - \int {\frac{{dx}}{x}} } \] ⇒ log |cosec v - cot v| = -logx + logc
log | cosec v - cot v| = log (c/x)
cosec v - cot v = c/x
\[x\left[ {\frac{1}{{\sin v}} - \frac{{\cos v}}{{\sin v}}} \right] = c\] \[x\left[ {1 - \cos \left( {\frac{y}{x}} \right)} \right] = c\sin \left( {\frac{y}{x}} \right)\]

Question (9)

\[ydx + x\log \left( {\frac{y}{x}} \right)dy - 2xdy = 0\]

Solution

\[\left[ {x\log \left( {\frac{y}{x}} \right) - 2x} \right]dy = - ydx\] \[\left[ {\log \left( {\frac{y}{x}} \right) - 2} \right]dy = - \frac{y}{x}dx\] \[\frac{{dy}}{{dx}} = \frac{{ - \frac{y}{x}}}{{\left[ {\log \left( {\frac{y}{x}} \right) - 2} \right]}}\] Let v=y/x
y= v x
differentiate w.r.t x \[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] \[ \therefore v + x\frac{{dv}}{{dx}} = \frac{{ - v}}{{\log v - 2}}\] \[x\frac{{dv}}{{dx}} = \frac{{v - v\log v}}{{\log v - 2}}\] \[\frac{{\left( {\log v - 2} \right)dv}}{{v - v\log v}} = \frac{{dx}}{x}\] \[\int {\frac{{\left( {\log v - 2} \right)dv}}{{v - v\log v}}} = \int {\frac{{dx}}{x}} \] \[ \Rightarrow \int {\frac{{\left( {\log v - 2} \right)dv}}{{v\left( {\log v - 1} \right)}}} = - \int {\frac{{dx}}{x}} \] let logv = t
∴ dv/v = dt \[\int {\frac{{t - 2}}{{t - 1}}dt = - \log x + \log c'} \] \[\int {\frac{{t - 1 - 1}}{{t - 1}}dt = \log \left( {\frac{c'}{x}} \right)} \] \[\int {dt} - \int {\frac{1}{{t - 1}}dt} = \log \left( {\frac{c'}{x}} \right)\] \[t - \log \left| {t - 1} \right| = \log \left( {\frac{c'}{x}} \right)\] \[\log v - \log \left| {\log \left( v \right) - 1} \right| = \log \frac{{c'}}{x}\] \[\log \left| {\frac{v}{{\log v - 1}}} \right| = \log \frac{{c'}}{x}\] \[ \Rightarrow \frac{v}{{\log v - 1}} = \frac{c'}{x}\] \[\log v - 1 = \frac{{vx}}{c'}\] \[\log \left| {\frac{y}{x}} \right| - 1 = cy\]

Question (10)

\[\left( {1 + {e^{\frac{x}{y}}}} \right)dx + {e^{\frac{x}{y}}}\left( {1 - \frac{x}{y}} \right)dy = 0\]

Solution

\[{e^{\frac{x}{y}}}\left( {1 - \frac{x}{y}} \right)dy = - \left( {1 + {e^{\frac{x}{y}}}} \right)dx\] \[\frac{{dx}}{{dy}} = \frac{{ - {e^{\frac{x}{y}}}\left( {1 - \frac{x}{y}} \right)}}{{\left( {1 + {e^{\frac{x}{y}}}} \right)}}\] let x/y = v
x = vy
differentiate w.r.t y
\[\frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}\] \[v + y\frac{{dv}}{{dy}} = \frac{{ - {e^v}\left( {1 - v} \right)}}{{1 + {e^v}}}\] \[y\frac{{dv}}{{dy}} = \frac{{ - {e^v}\left( {1 - v} \right)}}{{1 + {e^v}}} - v\] \[y\frac{{dv}}{{dy}} = \frac{{ - {e^v} + \cancel{v{e^v}} - v - \cancel{v{e^v}}}}{{1 + {e^v}}}\] \[\frac{{\left( {{e^v} + 1} \right)dv}}{{\left( {v + e^v} \right)}} = - \frac{{dy}}{y}\] \[\int {\frac{{\left( {{e^v} + 1} \right)dv}}{{\left( {v + e^v} \right)}}} = - \int {\frac{{dy}}{y}} \] let v+ev t
(1+ev) dv = dt
\[\int {\frac{{dt}}{t} = - \int {\frac{{dy}}{y}} } \] log t = -log y + log c
log t = log (c/y)
⇒ t = c/y
yt = c
y(v+ev) = c
vy + yev = c
\[\frac{x}{y} \cdot y + y{e^{\frac{x}{y}}} = c\] \[x + y{e^{\frac{x}{y}}} = c\] For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition

Question (11)

(x+y) dy + (x-y) dx = 0; y=1 when x = 1

Solution

(x+y) dy = -(x-y) dx
\[\frac{{dy}}{{dx}} = - \frac{{\cancel{x}\left( {1 - \frac{y}{x}} \right)}}{{\cancel{x}\left( {1 + \frac{y}{x}} \right)}}\] let y/x = v
y = v x
differentiate w.r.t x \[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] Replacing value \[v + x\frac{{dv}}{{dx}} = \frac{{ - \left( {1 - v} \right)}}{{1 + v}}\] \[x\frac{{dv}}{{dx}} = \frac{{ - \left( {1 - v} \right)}}{{1 + v}} - v\] \[x\frac{{dv}}{{dx}} = \frac{{ - 1 +\cancel{v} - v - {v^2}}}{{1 + v}}\] \[x\frac{{dv}}{{dx}} = \frac{{ - \left( {{v^2} + 1} \right)}}{{v + 1}}\] \[\frac{{v + 1}}{{{v^2} + 1}}dv = - \frac{{dx}}{x}\] \[\int {\frac{{v + 1}}{{{v^2} + 1}}dv} = - \int {\frac{{dx}}{x}} \] \[\frac{1}{2}\int {\frac{{2v}}{{{v^2} + 1}}dv + \int {\frac{1}{{{v^2} + 1}}dv = - \log x + c'} } \] \[\frac{1}{2}\log \left| {{v^2} + 1} \right| + {\tan ^{ - 1}}v = - \log x + c'\] log|v2+1| + 2tan-1v = -2logx + 2c' let 2c' =c \[\log \left| {\frac{{{y^2}}}{{{x^2}}} + 1} \right| + 2{\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = - 2\log x + c\] \[\log \left| {{y^2} + {x^2}} \right| - \cancel{2\log x} + 2{\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = - \cancel{2\log x} + c\] \[\log \left| {{y^2} + {x^2}} \right| + 2{\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = + c\] x=1 when y=1
log2+2tan-1 =c \[\log 2 + 2\frac{\pi }{4} = c\] \[\log 2 + \frac{\pi }{2} = c\] \[\log \left| {{y^2} + {x^2}} \right| + 2{\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = \log 2 + \frac{\pi }{2}\]

Question (12)

x2 dy + (xy+y2)dx = 0; y=1 when x = 1

Solution

x2 dy = - (xy+y2)dx \[\therefore \frac{{dy}}{{dx}} = - \frac{{\left( {xy + {y^2}} \right)}}{{{x^2}}}\] \[\frac{{dy}}{{dx}} = - \left[ {\frac{y}{x} + {{\left( {\frac{y}{x}} \right)}^2}} \right]\] let y/x = v
y = v x
differentiate w.r.t x \[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] \[v + x\frac{{dv}}{{dx}} = - \left[ {v + {v^2}} \right]\] \[x\frac{{dv}}{{dx}} = - v - {v^2} - v\] \[x\frac{{dv}}{{dx}} = - {v^2} - 2v\] \[\frac{{dv}}{{{v^2} + 2v}} = - \frac{{dx}}{x}\] \[\int {\frac{{dv}}{{{v^2} + 2v + 1 - 1}}} = - \int {\frac{{dx}}{x}} \] \[\int {\frac{{dv}}{{{{\left( {v + 1} \right)}^2} - {1^2}}}} = - \int {\frac{{dx}}{x}} \] \[\frac{1}{2}\log \left| {\frac{{v + 1 - 1}}{{v + 1 + 1}}} \right| = - \log x + \log c\] \[\log \left| {\frac{v}{{v + 2}}} \right| = \log \frac{c}{{{x^2}}}\] \[\frac{v}{{v + 2}} = \frac{c}{{{x^2}}}\] \[c\left( {v + 2} \right) = v{x^2}\] \[c\left( {\frac{y}{x} + 2} \right) = \frac{y}{\cancel{x}}{x^\cancel{2}}\] c(y+2x) = x2y
when x = 1, y = 1
c(3) = 1
c = 1/3
c(2x+y) = x2y
⅓ (2x+y) = x2y
2x + y = 3x2 y

Question (13)

\[\left[ {x{{\sin }^2}\left( {\frac{y}{x}} \right) - y} \right]dx + xdy = 0;\] y = π/4 when x=1

Solution

\[xdy = - \left[ {x{{\sin }^2}\left( {\frac{y}{x}} \right) - y} \right]dx\] \[\frac{{dy}}{{dx}} = - \left[ {{{\sin }^2}\left( {\frac{y}{x}} \right) - \frac{y}{x}} \right]dx\] Let y/x = v
y = vx
differentiate w.r.t x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] replace the value of dy/dx
\[v + x\frac{{dv}}{{dx}} = - {\sin ^2}v + v\] \[x\frac{{dv}}{{dx}} = - {\sin ^2}v\] \[\frac{{dv}}{{{{\sin }^2}v}} = - \frac{{dx}}{x}\] \[\int {{{cosec }^2}vdv = - \int {\frac{{dx}}{x}} } \] -cotv = -logx +c
cotv = logx - c
cot(y/x) = logx -c
x = 1 , y = π/4
cot(π/4) = log1 - c
- 1=c
cot (y/x) = log|x| + 1

Question (14)

\[\frac{{dy}}{{dx}} - \frac{y}{x} + cosec \left( {\frac{y}{x}} \right) = 0\] y= 0 when x = 1

Solution

let y/x = v
y = v x
differentiate w.r.t x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] \[\cancel{v} + x\frac{{dv}}{{dx}} - \cancel{v} + cosec v = 0\] \[x\frac{{dv}}{{dx}} = - {\mathop{\rm cosecv}\nolimits} \] \[\frac{{dv}}{{cosec v}} = - \frac{{dx}}{x}\] \[\int {\sin vdv = - \int {\frac{{dx}}{x}} } \] -cosv = -log x +logc
cosv=logx-logc
\[\cos v = \log \left| {\frac{x}{c}} \right|\] \[\cos \left( {\frac{y}{x}} \right) = \log \left| {\frac{x}{c}} \right|\] when x = 1, y=0
cos(o) = log|1/c|
1= log|1/c|
1/c = e
c = 1/e
\[\cos \left( {\frac{y}{x}} \right) = \log \left| {\frac{x}{{\frac{1}{e}}}} \right|\] \[\cos \left( {\frac{y}{x}} \right) = \log \left| {ex} \right|\]

Question (15)

\[2xy + {y^2} - 2{x^2}\frac{{dy}}{{dx}} = 0\] y =2 when x = 1

Solution

\[2xy + {y^2} = 2{x^2}\frac{{dy}}{{dx}}\] \[\frac{{dy}}{{dx}} = \frac{{2xy + {y^2}}}{{2{x^2}}}\] \[\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{1}{2}{\left( {\frac{y}{x}} \right)^2}\] let y/x = v
y = v x
differentiate w.r.t x
\[\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\] \[\cancel{v} + x\frac{{dv}}{{dx}} = \cancel{v} + \frac{{{v^2}}}{2}\] \[x\frac{{dv}}{{dx}} = \frac{{{v^2}}}{2}\] \[\frac{{2dv}}{{{v^2}}} = \frac{{dx}}{x}\] \[2\int {{v^{ - 2}}dv = \int {\frac{{dx}}{x}} } \] \[2\frac{{{v^{ - 1}}}}{{ - 1}} = \log x + c\] \[ - \frac{2}{v} = \log x + c\] \[ \Rightarrow - \frac{{2x}}{y} = \log x + c\] y = 2, x=1
\[\frac{{ - 2}}{2} = \log 1 + c\] c = -1
\[\frac{{ - 2x}}{y} = \log x - 1\] \[\frac{{2x}}{y} = 1 - \log x\] \[y = \frac{{2x}}{{1 - \log x}}\]

Question (16)

A homogeneous differential equation of the form \[\frac{{dx}}{{dy}} = h\left( {\frac{x}{y}} \right)\] can be solved by making the substitution (A) y = vx   (B) v =yx   (C) x = vy   (D) x = v

Solution

\[\frac{x}{y} = v \Rightarrow x = vy\] Correct option (C)

Question (17)

Which of the following is a homogeneous differential equation?
(A) (4x+6y+5)dy-(3y+2x+4)dx=0
(B) (xy)dx - (x3+y3)dy = 0
(C) (x3 +y2)dx + 2xy dy = 0
(D) y2 dx + (x2 - xy - y2) dy = 0

Solution

Homogeneous function
y2 dx + (x2 - xy - y2) dy = 0
\[\frac{{dy}}{{dx}} = \frac{{ - {y^2}}}{{{x^2} - xy + {y^2}}}\] Divide numerator and denominator by x2 \[\frac{{dy}}{{dx}} = \frac{{ - {{\left( {\frac{y}{x}} \right)}^2}}}{{1 - \left( {\frac{y}{x}} \right) + {{\left( {\frac{y}{x}} \right)}^2}}}\] so it is homogeneous
Option (D) is correct
Exercise 9.4⇐
⇒Exercise 9.6