Differential Equations MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Given, }}xdy = ydx + {y^2}dy \cr & \Rightarrow 1 = \frac{4}{x}\frac{{dx}}{{dy}} + \frac{{{y^2}}}{x} \cr & \Rightarrow \frac{{dx}}{{dy}} + y = \frac{x}{y} \cr & \Rightarrow \frac{{dx}}{{dy}} - \frac{x}{y} = - y......\left( 1 \right) \cr & P = - \frac{1}{y},\,Q = - y \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {Pdy} }} = {e^{\int { - \frac{1}{y}dy} }} = {e^{ - \log \,y}} = \frac{1}{y} \cr & {\text{Multiplying Equation }}\left( 1 \right){\text{ by I}}{\text{.F}}{\text{.}} \cr & \Rightarrow \frac{1}{y}\frac{{dx}}{{dy}} - \frac{x}{{{y^2}}} = - 1\,;\,\frac{x}{y} = \int {\frac{1}{y}\left( { - y} \right)dy + C} \cr & \Rightarrow \frac{x}{y} = \int { - 1\,dy + C} \cr & \Rightarrow \frac{x}{y} = - y + C \cr & \left( {y\left( 1 \right) = 1\, \Rightarrow \frac{1}{1} = - 1 + C\, \Rightarrow C = 2} \right) \cr & \Rightarrow \frac{x}{y} = - y + 2 \cr & \Rightarrow x = - {y^2} + 2y \cr & \Rightarrow y\left( { - 3} \right) \Rightarrow - 3 = - {y^2} + 2y \cr & \Rightarrow {y^2} - 2y - 3 = 0 \cr & \Rightarrow y = \frac{{ + 2 \pm \sqrt {4 + 12} }}{2} \cr & \Rightarrow y = \frac{{2 \pm 4}}{2} \cr & \Rightarrow y = 3,\,\, - 1 \cr & {\text{Since }}y > 0{\text{ so }}y = 3. \cr} $$

$${\left( {\frac{{dy}}{{dx}}} \right)^2} - x.\frac{{dy}}{{dx}} + y = 0$$
By actual verification we find that the choice (c),
i.e. $$y=2x-4$$   satisfies the given differential equation.

$$\frac{{dy}}{{dx}} = \frac{{x + y}}{x} = 1 + \frac{y}{x}$$
Putting $$y=vx$$   and $$\frac{{dv}}{{dx}} = v + x\frac{{dv}}{{dx}}$$
we get
$$\eqalign{ & v + x\frac{{dv}}{{dx}} = 1 + v\,\,\, \Rightarrow \int {\frac{{dx}}{x} = \int {dv} } \cr & \Rightarrow v = \ln \,x + c\,\, \Rightarrow y = x\,\ln \,x + cx \cr & {\text{As }}y\left( 1 \right) = 1 \cr & \therefore c = 1 \cr} $$
So solution is $$y=x \ln x + x$$

$$\eqalign{ & \frac{{dy}}{{dx}} = {e^x}\left( {A\cos \,x + B\sin \,x} \right) + {e^x}\left( { - A\sin \,x + B\cos \,x} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = y + {e^x}\left( { - A\sin \,x + B\cos \,x} \right) \cr & \therefore \frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy}}{{dx}} + {e^x}\left( { - A\sin \,x + B\cos \,x} \right) + {e^x}\left( { - A\cos \,x - B\sin \,x} \right) \cr & = \frac{{dy}}{{dx}} + \left( {\frac{{dy}}{{dx}} - y} \right) - y = 2\left( {\frac{{dy}}{{dx}} - y} \right) \cr & {\text{So, degree = }}1{\text{ and order}} = 2 \cr} $$

To eliminate the parameters $$g,\,f$$  and $$c$$ differentiate thrice w.r.t. $$x,$$
$$\eqalign{ & x + yy' + g + fy' = 0......\left( 1 \right) \cr & 1 + y{'^2} + yy'' + fy'' = 0......\left( 2 \right) \cr & 3y'y'' + yy''' + fy''' = 0......\left( 3 \right) \cr & \left( 1 \right)y''' - \left( 2 \right)y''{\text{ gives}} \cr & y''' + y{'^2}y''' - 3y'y'{'^2} = 0 \cr & \Rightarrow y''' = \frac{{3y'y'{'^2}}}{{1 + y{'^2}}} \cr} $$

Differential equation is
$$\left( {x + y} \right)\left( {dx - dy} \right) = dx + dy,$$       dividing by $$dx$$  on both the sides
$$\left( {x + y} \right)\left( {1 - \frac{{dy}}{{dx}}} \right) = 1 + \frac{{dy}}{{dx}}$$
Putting $$x + y = v$$
$$1 + \frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}{\text{ and }}\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} - 1$$
The equation changes to
$$\eqalign{ & v\left\{ {1 - \left( {\frac{{dv}}{{dx}} - 1} \right)} \right\} = \frac{{dv}}{{dx}}\,;\,v\left( {2 - \frac{{dv}}{{dx}}} \right) = \frac{{dv}}{{dx}} \cr & 2v - v\frac{{dv}}{{dx}} = \frac{{dv}}{{dx}}\,;\,2v = \left( {1 + v} \right)\frac{{dv}}{{dx}} \cr & \left( {\frac{{1 + v}}{v}} \right)dv = 2dx\,{\text{ or }}\left( {\frac{1}{v} + 1} \right)dv = 2dx \cr} $$
Integrating on both the sides
$$\eqalign{ & \int {\frac{{dv}}{v}} + \int {dv} = 2\int {dx + c} \cr & {\text{or, }}\log \,v + v = 2x + c\,\,\,\,\,\,\,\,\,\left[ {{\text{Putting }}v = x + y} \right] \cr & {\text{or, }}\log \left( {x + y} \right) + x + y = 2x + c \cr & {\text{or, }}\log \left( {x + y} \right) + y - x = c \cr & {\text{or, }}y - x + \log \left( {x + y} \right) = c \cr} $$

Let the initial population be $${x_0}$$ and it is $$x$$ in $$t$$ years, then the differential equation is
$$\frac{{dx}}{{dt}} = kx,\,\,k$$   is a constant
$$ \Rightarrow \,\frac{{dx}}{x} = k\,dt$$
Integrating we get
$$\log \,x + kt + c......\left( {\text{i}} \right)$$
When $$t = 0,\,x = {x_0} \Rightarrow c = \log \,{x_0}$$
Then from equation $$\left( {\text{i}} \right),$$
$$\eqalign{ & \log \,x = kt + \log \,{x_0} \cr & \Rightarrow \log \frac{x}{{{x_0}}} = kt......\left( {{\text{ii}}} \right) \cr} $$
Now when $$t = 40,\,\frac{x}{{{x_0}}} = 2$$
$$\eqalign{ & \Rightarrow \log \,2 = k.40 \cr & \Rightarrow k = \frac{{\log \,2}}{{40}} \cr} $$
$$\eqalign{ & \therefore \,{\text{equation }}\left( {{\text{ii}}} \right){\text{ bocomes }}\log \frac{x}{{{x_0}}} = \frac{{\log \,2}}{{40}}.t \cr & {\text{Next put }}\frac{x}{{{x_0}}} = 3 \Rightarrow t = 40\frac{{\log \,3}}{{\log \,2}} \cr} $$

Given differential equation is
$$\frac{{dp\left( t \right)}}{{dx}} = \frac{1}{2}p\left( t \right) - 200$$
By separating the variable, we get
$$\eqalign{ & dp\left( t \right) = \left[ {\frac{1}{2}p\left( t \right) - 200} \right]dt \cr & \Rightarrow \frac{{dp\left( t \right)}}{{\frac{1}{2}p\left( t \right) - 200}} = dt \cr} $$
Integrating on both the sides,
$$\eqalign{ & \int {\frac{{d\left( {p\left( t \right)} \right)}}{{\frac{1}{2}p\left( t \right) - 200}}} = \int {dt} \cr & {\text{Let }}\frac{1}{2}p\left( t \right) - 200 = s \cr & \Rightarrow \frac{{dp\left( t \right)}}{2} = ds \cr & {\text{So, }}\int {\frac{{dp\left( t \right)}}{{\frac{1}{2}p\left( t \right) - 200}}} = \int {dt} \cr & \Rightarrow \int {\frac{{2ds}}{s} = \int {dt} } \cr & \Rightarrow 2\,\log \,s = t + c \cr & \Rightarrow 2\,\log \left( {\frac{{p\left( t \right)}}{2} - 200} \right) = t + c \cr & \Rightarrow \frac{{p\left( t \right)}}{2} - 200 = {e^{\frac{1}{2}}}k \cr} $$
Using given condition $$p\left( t \right) = 400 - 300\,{e^{\frac{t}{2}}}$$

$${y^2} = 4a\left( {x - h} \right),\,\,2y{y_1} = 4a\,\, \Rightarrow y{y_1} = 2a$$
Differentiating, $$ \Rightarrow y_1^2 + y{y_2} = 0$$
Degree $$= 1,$$   order $$=2.$$

$$\eqalign{ & \frac{{dy}}{{dx}}\left( {\frac{{2 + \sin \,x}}{{1 + y}}} \right) = - \cos \,x,\,y\left( 0 \right) = 1 \cr & \Rightarrow \frac{{dy}}{{\left( {1 + y} \right)}} = \frac{{ - \cos \,x}}{{2 + \sin \,x}}dx \cr & {\text{Integrating both sides}} \cr & \Rightarrow \ln \left( {1 + y} \right) = - \ln \left( {2 + \sin \,x} \right) + C \cr & {\text{Put}}\,x = 0\,{\text{and}}\,y = 1 \cr & \Rightarrow \ln \left( 2 \right) = - \ln \,2 + C \cr & \Rightarrow C = \ln \,4 \cr & {\text{Put }}x = \frac{\pi }{2} \cr & \ln \left( {1 + y} \right) = - \ln \,3 + \ln \,4 \cr & \Rightarrow \ln \left( {1 + y} \right) = \ln \frac{4}{3} \cr & \Rightarrow y = \frac{1}{3} \cr} $$