Differential Equations MCQ Questions & Answers in Calculus | Maths

The differential equation is
$$\eqalign{ & \frac{{dy}}{{dx}} - \frac{y}{x} = - \frac{{5x}}{{\left( {x + 2} \right)\left( {x - 3} \right)}} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\left( { - \frac{1}{x}} \right)dx} }} = {e^{ - \ln \,x}} = \frac{1}{x} \cr} $$
Solution is
$$y\left( {\frac{1}{x}} \right) = \int {\left( {\frac{1}{x}} \right) \times \frac{{5x}}{{\left( {x + 2} \right)\left( {x - 3} \right)}}dx} = \ln \left( {\frac{{x + 2}}{{x - 3}}} \right) + C$$
It passes through $$\left( {4,\,0} \right),\,{\text{so }}C = - \ln \,6$$
$$\eqalign{ & \therefore \,y = x\,\ln \left\{ {\frac{{\left( {x + 2} \right)}}{{6\left( {x - 3} \right)}}} \right\}{\text{ }}\,{\text{Putting }}\left( {5,\,a} \right) \cr & {\text{we get }}a = 5\,\ln \,\left( {\frac{7}{{12}}} \right) \cr} $$

Given : $$\frac{{dy}}{{dx}} = \frac{{ax + b}}{{cy + d}}$$
or, $$\left( {cy + d} \right)dy = \left( {ax + b} \right)dx$$
Integrating both the sides.
$$\eqalign{ & c.\int {y\,dy + d} \int {dy = a} \int {x\,dx + b} \int {dx + K} \,\,\,\,\,\,\left[ {K{\text{ is constant integration}}} \right] \cr & {\text{or, }}c.\frac{{{y^2}}}{2} + d.y = a\frac{{{x^2}}}{2} + b.x + K \cr & {\text{or, }}c{y^2} + 2d.y = a{x^2} + 2b.x + 2K \cr} $$
This equation will represent a parabola when either, the coefficient of $${x^2}$$ or the coefficient of $${y^2}$$ is zero, but not both.
Thus either $$c = 0$$  or $$a = 0$$  but not both.
From the choice given $$a = 0,\,c \ne 0{\text{ and }}\,b \ne 0$$

Putting $$x + y + 1 = u,$$    we have $$du = dx + dy$$   and the given equation reduces to $$u\left( {du - dx} \right) = dx$$
$$\eqalign{ & \Rightarrow \frac{{u\,du}}{{u + 1}} = dx \cr & \Rightarrow u - \log \left( {u + 1} \right) = x \cr & \Rightarrow \log \left( {x + y + 2} \right) = y + {\text{constant}} \cr & \Rightarrow x + y + 2 = C{e^y} \cr} $$

$$\eqalign{ & {\text{We have, }}dy + \left\{ {y\phi '\left( x \right) - \phi \left( x \right)\phi '\left( x \right)} \right\}dx = 0 \cr & \Rightarrow \frac{{dy}}{{dx}} + \phi '\left( x \right).y = \phi \left( x \right)\phi '\left( x \right)......\left( {\text{i}} \right) \cr} $$
This is a linear differential equation with $${\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\phi '\left( x \right)} dx}} = {e^{\phi \left( x \right)}}$$
Multiplying Equation $$\left( {\text{i}} \right)$$ by $$\phi \left( x \right)$$  and integrating, we get
$$\eqalign{ & y{e^{\phi \left( x \right)}} = \int {\phi \left( x \right)\phi '\left( x \right){e^{\phi \left( x \right)}}dx} \cr & \Rightarrow y{e^{\phi \left( x \right)}} = \int {{e^{\phi \left( x \right)}}\phi \left( x \right)\phi '\left( x \right)dx} \cr & \Rightarrow y{e^{\phi \left( x \right)}} = \int {\phi \left( x \right){e^{\phi \left( x \right)}}\phi '\left( x \right)dx} \cr & \Rightarrow y{e^{\phi \left( x \right)}} = \phi \left( x \right){e^{\phi \left( x \right)}} - \int {\phi '\left( x \right){e^{\phi \left( x \right)}}dx} \cr & \Rightarrow y{e^{\phi \left( x \right)}} = \phi \left( x \right){e^{\phi \left( x \right)}} - {e^{\phi \left( x \right)}} + c \cr & \Rightarrow y = \left\{ {\phi \left( x \right) - 1} \right\} + c{e^{ - \phi \left( x \right)}} \cr} $$

$$\eqalign{ & y = 2\,\cos \,x + 3\,\sin \,x \cr & \frac{{dy}}{{dx}} = - 2\,\sin \,x + 3\,\cos \,x\,; \cr & \frac{{{d^2}y}}{{d{x^2}}} = - 2\,\cos \,x - 3\,\sin \,x \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \left( {2\,\cos \,x + 3\,\sin \,x} \right) \cr & \frac{{{d^2}y}}{{d{x^2}}} = - y\,; \cr & \frac{{{d^2}y}}{{d{x^2}}} + y = 0 \cr} $$

The equation is
$$x\int_0^x {y\left( t \right)dt = \left( {x + 1} \right)} \int_0^x {ty\left( t \right)dt......\left( {\text{i}} \right)} $$
Differentiating both the sides with respect to $$x,$$ we get
$$\eqalign{ & xy\left( x \right) + \int_0^x {y\left( t \right)dt = \left( {x + 1} \right)xy\left( x \right)} + \int_0^x {ty\left( t \right)dt} \cr & \Rightarrow \int_0^x {y\left( t \right)dt = {x^2}y\left( x \right)} + \int_0^x {ty\left( t \right)dt......\left( {{\text{ii}}} \right)} \cr} $$
Differentiating again with respect to $$x,$$ we get
$$\eqalign{ & y\left( x \right) = 2xy\left( x \right) + {x^2}y'\left( x \right) + xy\left( x \right) \cr & \Rightarrow {x^2}\frac{{dy}}{{dx}} = \left( {1 - 3x} \right)y\,\,\,\,\,\,\,\,\left[ {{\text{writing }}y\left( x \right) = y} \right] \cr & \Rightarrow \frac{{dy}}{y} = \left( {\frac{{1 - 3x}}{{{x^2}}}} \right)dx \cr & \Rightarrow \frac{{dy}}{y} = \left( {\frac{1}{{{x^2}}} - \frac{3}{x}} \right)dx \cr} $$
Integrating we get, $$\log \,y = - \frac{1}{x} - 3\,\log \,x + a,\,\,\,a$$       is constant
$$\eqalign{ & \Rightarrow \log \,y + 3\,\log \,x = a - \frac{1}{x} \cr & \Rightarrow \log \left( {y{x^3}} \right) = a - \frac{1}{x} \cr & \Rightarrow y{x^3} = {e^{a - \frac{1}{x}}} \cr & \Rightarrow y{x^3} = c.{e^{ - \frac{1}{x}}}{\text{ where }}c = {e^a} \cr & \therefore \,y = \frac{c}{{{x^3}}}{e^{ - \frac{1}{x}}} \cr} $$

In the given equation $$K.\frac{{{d^2}y}}{{d{x^2}}} = {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^3}} \right]^{\frac{3}{2}}}$$
Squaring both the sides $${K^2}.{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} = {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^3}} \right]^3}$$
Degree of a differential equation is the highest power of the highest derivative in equation when derivatives are expressed as polynomial. Here degree of differential equation is $$2.$$

The family of curves which are orthogonal (i.e. intersect at right angles) to a given system of curves is obtained by substituting $$ - \frac{{dx}}{{dy}}$$  for $$\frac{{dy}}{{dx}}$$  in the differential equation of the given system.
The given differential equation is $${\left( {\frac{{dy}}{{dx}}} \right)^2} = \frac{a}{x}$$
Replacing $$\frac{{dy}}{{dx}}$$  by $$ - \frac{{dx}}{{dy}},$$  we get
$${\left( {\frac{{dx}}{{dy}}} \right)^2} = \frac{a}{x}\, \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)^2} = \frac{x}{a}\, \Rightarrow \frac{{dy}}{{dx}} = \pm \sqrt {\frac{x}{a}} $$
Integrating we get
$$\eqalign{ & y + c = \pm \frac{2}{{3\sqrt a }}{x^{\frac{3}{2}}}......\left( {\text{i}} \right) \cr & \Rightarrow {\left( {y + c} \right)^2} = \frac{4}{{9a}}{x^3} \cr & \Rightarrow 9a{\left( {y + c} \right)^2} = 4{x^3}......\left( {{\text{ii}}} \right) \cr} $$
From $$\left( {{\text{i}}} \right)$$ and $$\left( {{\text{ii}}} \right)$$ all of the first three given options represent required equations.

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

Given $${x^2} + {y^2} = 1.$$
Differentiating w.r.t. $$x,$$ we get $$2x + 2yy' = 0\,\,{\text{or }}x + yy' = 0$$
Again differentiating w.r.t. $$x,$$ we get $$1 + y'y' + yy'' = 0\,\,{\text{or 1}} + {\left( {y'} \right)^2} + yy'' = 0$$