Differential Equations MCQ Questions & Answers in Calculus | Maths

$$y = mx + c$$    (Equation of straight line)
$$\frac{{dy}}{{dx}} = m$$   and $$mx - y + c = 0$$    is at unit distance from origin.
$$\eqalign{ & \therefore \,\frac{{\left| {m\left( 0 \right) - \left( 0 \right) + c} \right|}}{{\sqrt {{m^2} + {{\left( { - 1} \right)}^2}} }} = 1\, \Rightarrow c = \sqrt {1 + {m^2}} \cr & {\text{Now}}\,{\text{:}}\,{\left[ {y - x\frac{{dy}}{{dx}}} \right]^2} = {\left[ {mx + c - xm} \right]^2} = {c^2} = 1 + {m^2} \cr & {\text{also,}}\,{\left[ {y + x\frac{{dy}}{{dx}}} \right]^2} = {\left[ {mx + c + xm} \right]^2} = {\left[ {2mx + \sqrt {1 + {m^2}} } \right]^2} \cr & {\text{also,}}\,1 - {\left( {\frac{{dy}}{{dx}}} \right)^2} = 1 - {m^2}{\text{ and }}1 + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 1 + {m^2} \cr & \Rightarrow {\left[ {y - x\frac{{dy}}{{dx}}} \right]^2} = 1 + {\left( {\frac{{dy}}{{dx}}} \right)^2} \cr} $$

Given differential equation is $$y = x\frac{{dy}}{{dx}} + {\left( {\frac{{dy}}{{dx}}} \right)^{ - 1}}$$
Multiply by $$\frac{{dy}}{{dx}}\,;\,\,\,y\frac{{dy}}{{dx}} = x{\left( {\frac{{dy}}{{dx}}} \right)^2} + 1$$
Since power of highest order derivative is $$2$$
$$\therefore $$  degree $$ = 2.$$

The given D.E. is $$\left( {{x^2} + {y^2}} \right)dy = xy\,dx\,\,\,{\text{s}}{\text{.t}}{\text{.}}\,\,y\left( 1 \right) = 1$$       and $$y\left( {{x_0}} \right) = e$$
The given equation can be written as
$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}} \cr & {\text{Put }}y = vx, \cr & \therefore v + x\frac{{dv}}{{dx}} = \frac{v}{{1 + {v^2}}} \cr & \Rightarrow x\frac{{dv}}{{dx}} = \frac{{ - {v^3}}}{{1 + {v^2}}} \cr & \Rightarrow \int {\frac{{1 + {v^2}}}{{{v^3}}}dv + \int {\frac{{dx}}{x} = 0} } \cr & \Rightarrow - \frac{1}{{2{v^2}}} + \log \left| v \right| + \log \left| x \right| = C \cr & \Rightarrow \log \,y = C + \frac{{{x^2}}}{{2{y^2}}}\,\,\,\,\,\,\,\left( {{\text{using }}v = \frac{y}{x}} \right) \cr & {\text{Also, }}y\left( 1 \right) = 1 \Rightarrow \log 1 = C + \frac{1}{2} \Rightarrow C = - \frac{1}{2} \cr & \therefore \,\log \,y = \frac{{{x^2} - {y^2}}}{{2{y^2}}},\,\,\,\,\,\,\,\,{\text{But given }}y\left( {{x_0}} \right) = e \cr & \Rightarrow \log \,e = \frac{{x_0^2 - {e^2}}}{{2{e^2}}}\,\,\,\, \Rightarrow x_0^2 = 3{e^2}\,\,\,\, \Rightarrow {x_0} = \sqrt 3 \,e \cr} $$

$$\eqalign{ & {\text{We have }}\frac{{dy}}{{dx}} = \frac{{f'\left( x \right)}}{{f\left( x \right)}}y - \frac{{{y^2}}}{{f\left( x \right)}} \cr & \Rightarrow \frac{{dy}}{{dx}} - \frac{{f'\left( x \right)}}{{f\left( x \right)}}y = - \frac{{{y^2}}}{{f\left( x \right)}} \cr & {\text{Divide by }}{y^2} \cr & {y^{ - 2}}\frac{{dy}}{{dx}} - {y^{ - 1}}\frac{{f'\left( x \right)}}{{f\left( x \right)}} = - \frac{1}{{f\left( x \right)}} \cr & {\text{Put }}{y^{ - 1}} = z \cr & \Rightarrow - {y^{ - 2}}\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}} - \frac{{dz}}{{dx}} - \frac{{f'\left( x \right)}}{{f\left( x \right)}}\left( z \right) = - \frac{1}{{f\left( x \right)}} \cr & \Rightarrow \frac{{dz}}{{dx}} + \frac{{f'\left( x \right)}}{{f\left( x \right)}}\left( z \right) = \frac{1}{{f\left( x \right)}} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx} }} = {e^{\log \,f\left( x \right)}} = f\left( x \right) \cr & \therefore {\text{The solution is }}z\left( {f\left( x \right)} \right) = \int {\frac{1}{{f\left( x \right)}}\left( {f\left( x \right)} \right)dx + c} \cr & \Rightarrow {y^{ - 1}}\left( {f\left( x \right)} \right) = x + c \cr & \Rightarrow f\left( x \right) = y\left( {x + c} \right) \cr} $$

$$\eqalign{ & \because \,\frac{{dy}}{{dx}} - x = {\left( {y - x\frac{{dy}}{{dx}}} \right)^{ - 4}} \cr & \Rightarrow \left( {\frac{{dy}}{{dx}} - x} \right){\left( {y - x.\frac{{dy}}{{dx}}} \right)^4} = 1 \cr} $$
$$\therefore $$  Order of the above differential equation $$= 1$$  & degree $$= 5$$

$$\eqalign{ & \left( {1 + {y^2}} \right)\frac{{dx}}{{dy}} + x = {e^{{{\tan }^{ - 1}}y}} \cr & {\text{or }}\,\frac{{dx}}{{dy}} + \frac{1}{{1 + {y^2}}}.x = \frac{{{e^{{{\tan }^{ - 1}}y}}}}{{1 + {y^2}}}\,\,{\text{which is in the linear form}}{\text{.}} \cr & {\text{IF}} = {e^{\int {\frac{1}{{1 + {y^2}}}dy} }} = {e^{{{\tan }^{ - 1}}y}}.\,\,\,{\text{So, }}\,x.{e^{{{\tan }^{ - 1}}y}} = \int {\frac{{\left( {{e^{{{\tan }^{ - 1}}{y^2}}}} \right)dy}}{{1 + {y^2}}}} \cr & \therefore x{e^{{{\tan }^{ - 1}}y}} = \frac{{{{\left( {{e^{{{\tan }^{ - 1}}y}}} \right)}^2}}}{2} + \frac{k}{2},\,\,\,\,\left( {{\text{putting }}{e^{{{\tan }^{ - 1}}y}} = z} \right) \cr} $$

$$\eqalign{ & \frac{{dx}}{{dy}} + \int {y.\,dx} = {x^3} \cr & \Rightarrow \int {y.dx} = {x^3} - \frac{{dx}}{{dy}} \cr & \Rightarrow 1 + \frac{{dy}}{{dx}}\left( {\int {y.dx} } \right) = {x^3}.\frac{{dy}}{{dx}} \cr} $$
Differentiate both sides w.e.t. $$x$$
$$\eqalign{ & \Rightarrow 0 + \frac{{dy}}{{dx}}\left( y \right) + \left( {\int {y.dx} } \right)\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) = {x^3}.\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}}\left( {2{x^2}} \right) \cr & \Rightarrow y.\frac{{dy}}{{dx}} + \frac{{{d^2}y}}{{d{x^2}}}\left[ {{x^3} - \frac{{dx}}{{dy}}} \right] = {x^3}.\frac{{{d^2}y}}{{d{x^2}}} + 2{x^2}\frac{{dy}}{{dx}} \cr & \Rightarrow y\frac{{dy}}{{dx}} + {x^3}\frac{{{d^2}y}}{{d{x^2}}} - \left( {\frac{{dx}}{{dy}}} \right)\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) = {x^3}.\frac{{{d^2}y}}{{d{x^2}}} + 2{x^2}\frac{{dy}}{{dx}} \cr & \Rightarrow y\frac{{dy}}{{dx}} - \frac{{dx}}{{dy}}.\frac{{{d^2}y}}{{d{x^2}}} = 2{x^2}.\frac{{dy}}{{dx}} \cr} $$
Multiplying both side by $$\frac{{dy}}{{dx}}$$
$$\eqalign{ & y{\left( {\frac{{dy}}{{dx}}} \right)^2} - \frac{{{d^2}y}}{{d{x^2}}} = 2{x^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} \cr & \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} + \left( {2{x^2} - y} \right){\left( {\frac{{dy}}{{dx}}} \right)^2} = 0 \cr & {\text{Order }} = 2,\,\,{\text{degree }} = 1 \cr} $$

$$\frac{{{d^2}y}}{{d{x^2}}} = {e^{ - 2x}}\,\,;\,\,\frac{{dy}}{{dx}} = \frac{{{e^{ - 2x}}}}{{ - 2}} + c\,\,;\,\,y = \frac{{{e^{ - 2x}}}}{4} + cx + d$$

Given, Rate of change is
$$\eqalign{ & \frac{{dP}}{{dx}} = 100 - 12\sqrt x \cr & \Rightarrow dP = \left( {100 - 12\sqrt x } \right)dx \cr} $$
By integrating
$$\eqalign{ & \Rightarrow \int {dP = \int {\left( {100 - 12\sqrt x } \right)dx} } \cr & P = 100x - 8{x^{\frac{3}{2}}} + C \cr} $$
Given, when $$x=0$$   then $$P =2000$$
$$ \Rightarrow C = 2000$$
Now when $$x= 25$$  then
$$\eqalign{ & P = 100 \times 25 - 8 \times {\left( {25} \right)^{\frac{3}{2}}} + 2000 \cr & \Rightarrow P = 4500 - 1000 \cr & \Rightarrow P = 3500 \cr} $$

The equation is $$\frac{{dy}}{{dx}} + \left( {\frac{{x\,\cos \,x + \sin \,x}}{{x\,\sin \,x}}} \right)y = \frac{1}{x}$$
Integrating factor
$${\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{{x\,\cos \,x + \sin \,x}}{{x\,\sin \,x}}} dx}} = {e^{\log \left( {x\,\sin \,x} \right)}} = x\,\sin \,x$$
$$\therefore $$  The solution is
$$\eqalign{ & y\left( {x\,\sin \,x} \right) = \int {\frac{1}{x}\left( {x\,\sin \,x} \right)dx + c} \cr & xy\,\sin \,x = - \cos \,x + c{\text{ when}} \cr & x = 0,\,y = 0 \Rightarrow c = \cos \,0 = 1 \cr} $$
$$\therefore $$  The particular solution is $$xy\,\sin \,x = 1 - \cos \,x$$