Differential Equations MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {y^2} = 2c\left( {x + \sqrt c } \right)\,.....({\text{i}}) \cr & 2yy' = 2c.1\,\,{\text{or}}\,\,yy' = c\,.....({\text{ii}}) \cr & \Rightarrow {y^2} = 2yy'\left( {x + \sqrt {yy'} } \right) \cr} $$
[On putting value of $$c$$ from (ii) in (i)]
On simplifying, we get
$${\left( {y - 2xy'} \right)^2} = 4yy{'^3}\,.....({\text{iii}})$$
Hence equation (iii) is of order 1 and degree 3

Rearranging the equation, we have
$$\eqalign{ & dx - y\,dy + \sqrt {\left( {{x^2} + {y^2}} \right)} \left( {x\,dx + y\,dy} \right) = 0 \cr & \Rightarrow dx - y\,dy + \frac{1}{2}\sqrt {\left( {{x^2} + {y^2}} \right)} d\left( {{x^2} + {y^2}} \right) = 0 \cr} $$
On integrating, we get
$$\eqalign{ & x - \frac{{{y^2}}}{2} + \frac{1}{2}\int {\sqrt t \,dt = c,\,\left\{ {t = \sqrt {\left( {{x^2} + {y^2}} \right)} } \right\}} \cr & {\text{or }}x - \frac{{{y^2}}}{2} + \frac{1}{3}{\left( {{x^2} + {y^2}} \right)^{\frac{3}{2}}} = C \cr} $$

$$\eqalign{ & {\text{We have, }}\left( {{x^2} + {y^2}} \right)dy = xy\,dx \cr & \Rightarrow \frac{{dy}}{{dx}} = \frac{{xy}}{{{x^2} + {y^2}}} \cr & {\text{Substitute }}y = vx \Rightarrow \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}} \cr & \Rightarrow v + x\frac{{dv}}{{dx}} = \frac{v}{{1 + {v^2}}} \cr & \Rightarrow x\frac{{dv}}{{dx}} = \frac{v}{{1 + {v^2}}} - v = - \frac{{{v^3}}}{{1 + {v^2}}} \cr & \Rightarrow \frac{{1 + {v^2}}}{{{v^3}}}dv = - \frac{{dx}}{x} \cr} $$
Integrating both sides we get, $$ - \frac{1}{{2{v^2}}} + \log \,v = - \log \,x + \log \,c,$$       where $$c$$ is constant
$$\eqalign{ & \Rightarrow - \frac{1}{{2{v^2}}} + \log \left( {\frac{{vx}}{c}} \right) = 0 \cr & \Rightarrow y = c{e^{\frac{{{x^2}}}{{2{y^2}}}}} \cr & {\text{Now using }}y\left( 1 \right) = 1 \Rightarrow c = {e^{ - \frac{1}{2}}} \cr & {\text{Also, }}y\left( {{x_2}} \right) = e \Rightarrow e = {e^{ - \frac{1}{2} + \frac{{x_0^2}}{{2{e^2}}}}} \cr & \Rightarrow 1 = - \frac{1}{2} + \frac{{x_0^2}}{{2{e^2}}} \cr & \Rightarrow x_0^2 = 3{e^2} \cr & \Rightarrow {x_0} = \sqrt 3 e \cr} $$

The differential equation is $$\frac{{{d^2}y}}{{d{x^2}}} = 1 + \sin \,x.....\left( {\text{i}} \right)$$
Integrating we get $$\frac{{dy}}{{dx}} = x - \cos \,x + c......\left( {{\text{ii}}} \right)$$
When $$x = 0,\,\frac{{dy}}{{dx}} = 0 \Rightarrow c = 1$$
$$\therefore $$  Equation $$\left( {{\text{ii}}} \right)$$ is $$\frac{{dy}}{{dx}} = x - \cos \,x + 1$$
Integrating again we get $$y = \frac{{{x^2}}}{2} - \sin \,x + x + D......\left( {{\text{iii}}} \right)$$
When $$x = 0,\,y = 0\, \Rightarrow D = 0$$
$$\therefore $$  The particular solution is $$y = \frac{{{x^2}}}{2} + x - \sin \,x$$

The equation of the family is $$\frac{{{x^2}}}{{{a^2} + \lambda }} + \frac{{{y^2}}}{{{b^2} + \lambda }} = 1$$     because they have the same foci $$\left( { \pm \sqrt {{a^2} + {b^2}} ,\,0} \right).$$
On differentiation,
$$\eqalign{ & \frac{{2x}}{{{d^2} + \lambda }} + \frac{{2y}}{{{b^2} + \lambda }}.\frac{{dy}}{{dx}} = 0 \cr & {\text{or }}\frac{x}{{{a^2} + \lambda }} + \frac{{yp}}{{{b^2} + \lambda }} = 0,\,\,\left( {p = \frac{{dy}}{{dx}}} \right) \cr & {\text{or }}x\left( {{b^2} + \lambda } \right) + yp\left( {{a^2} + \lambda } \right) = 0 \cr & \Rightarrow \lambda = \frac{{ - {b^2}x - {a^2}yp}}{{x + yp}} \cr} $$
$$\therefore $$ the differential equation is,
$$\eqalign{ & \frac{{{x^2}}}{{{a^2} + \frac{{ - {b^2}x - {a^2}yp}}{{x + yp}}}} + \frac{{{y^2}}}{{{b^2} + \frac{{ - {b^2}x - {a^2}yp}}{{x + yp}}}} = 1 \cr & {\text{or }}\frac{{x\left( {x + yp} \right)}}{{{a^2} - {b^2}}} + \frac{{y\left( {x + yp} \right)}}{{\left( {{b^2} - {a^2}} \right)p}} = 1 \cr} $$
So, order $$=1$$   and degree $$=2.$$

Given that $${y^2} = p\left( x \right)$$
Differentiating
$$\eqalign{ & \Rightarrow 2y{y_1} = p'\left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{here }}{y_1} = \frac{{dy}}{{dx}}} \right] \cr & \Rightarrow 2{y_1} = \frac{{p'\left( x \right)}}{y} \cr} $$
Differentiating again,
$$\eqalign{ & \Rightarrow 2{y_2} = \frac{{yp''\left( x \right) - p'\left( x \right){y_1}}}{{{y^2}}},\,\,\,\,\,\,\,\,\,\,\left[ {{\text{here }}{y_2} = \frac{{{d^2}y}}{{d{x^2}}}} \right] \cr & \Rightarrow 2{y_2} = \frac{{yp''\left( x \right) - \frac{{p'\left( x \right).p'\left( x \right)}}{{2y}}}}{{{y^2}}} \cr & \Rightarrow 2{y_2} = \frac{{2{y^2}p''\left( x \right) - {{\left( {p'\left( x \right)} \right)}^2}}}{{{2y^3}}} \cr & \Rightarrow 2{y^3}{y_2} = \frac{1}{2}\left[ {2{y^2}p''\left( x \right) - {{\left( {p'\left( x \right)} \right)}^2}} \right] \cr & \Rightarrow 2{y^3}{y_2} = \frac{1}{2}\left[ {2p\left( x \right)p''\left( x \right) - {{\left( {p'\left( x \right)} \right)}^2}} \right] \cr & \Rightarrow 2\frac{d}{{dx}}\left( {{y^3}{y_2}} \right) = \frac{1}{2}\left[ {2p'\left( x \right) + 2p\left( x \right)p'''\left( x \right) - 2p'\left( x \right)p''\left( x \right)} \right] \cr & \Rightarrow 2\frac{d}{{dx}}\left( {{y^3}{y_2}} \right) = p\left( x \right)p'''\left( x \right) \cr} $$

Any conic whose axes coincide with coordinate axis is
$$a{x^2} + b{y^2} = 1......\left( {\text{i}} \right)$$
Diff. both sides w.r.t. $$'x',$$ we get
$$2ax + 2by\frac{{dy}}{{dx}} = 0{\text{ i}}{\text{.e}}{\text{., }}ax + by\frac{{dy}}{{dx}} = 0......\left( {{\text{ii}}} \right)$$
Diff. again $$a + b\left( {y\frac{{{d^2}y}}{{d{x^2}}} + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right) = 0......\left( {{\text{iii}}} \right)$$
From $$\left( {{\text{ii}}} \right),\,\,\,\frac{a}{b} = - \frac{{y\frac{{\,dy}}{{dx}}}}{x}$$
From $$\left( {{\text{iii}}} \right),\,\,\frac{a}{b} = - \,\left( {y\frac{{{d^2}y}}{{d{x^2}}} + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)$$
$$\eqalign{ & \therefore \,\frac{{y\frac{{dy}}{{dx}}}}{x} = y\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} \cr & \Rightarrow xy\frac{{{d^2}y}}{{d{x^2}}} + x{\left( {\frac{{dy}}{{dx}}} \right)^2} - y\frac{{dy}}{{dx}} = 0 \cr} $$

Divide the equation by $$\sec \,y$$
$$\eqalign{ & \cos \,y\frac{{dy}}{{dx}} - \frac{{\sin \,y}}{{1 + x}} = \left( {1 + x} \right){e^x} \cr & {\text{Put }}\sin \,y = z \Rightarrow \cos \,y\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}{\text{ then}} \cr & \frac{{dz}}{{dx}} - \left( {\frac{1}{{1 + x}}} \right)z = \left( {1 + x} \right){e^x} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{ - \int {\frac{1}{{1 + x}}dx} }} = {e^{ - \log \left( {1 + x} \right)}} = \frac{1}{{1 + x}} \cr} $$
The solution is $$z\left( {\frac{1}{{1 + x}}} \right) = {e^x} + c \Rightarrow \sin \,y = \left( {1 + x} \right)\left( {{e^x} + c} \right)$$

The intersection of $$y - x + 1 = 0$$    and $$y + x + 5 = 0$$    is $$\left( { - 2,\, - 3} \right).$$
Put $$x = X - 2,\,y = Y - 3.$$
The given equation reduces to $$\frac{{dY}}{{dX}} = \frac{{Y - X}}{{Y + X}}.$$
This is a homogeneous equation.
Putting $$Y = vX,$$   we get
$$\eqalign{ & X\frac{{dv}}{{dX}} = - \frac{{{v^2} + 1}}{{v + 1}} \cr & \Rightarrow \left( { - \frac{v}{{{v^2} + 1}} - \frac{1}{{{v^2} + 1}}} \right)dv = \frac{{dX}}{X} \cr & \Rightarrow - \frac{1}{2}\log \left( {{v^2} + 1} \right) - {\tan ^{ - 1}}v = \log \left| X \right| + {\text{constant}} \cr & \Rightarrow \log \left( {{Y^2} + {X^2}} \right) + 2\,{\tan ^{ - 1}}\frac{Y}{X} = {\text{constant}} \cr & \Rightarrow \log \left( {{{\left( {y + 3} \right)}^2} + {{\left( {x + 2} \right)}^2}} \right) + 2\,{\tan ^{ - 1}}\frac{{y + 3}}{{x + 2}} = C \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} + 2y = 1\,;\,\frac{{dy}}{{dx}} = 1 - 2y \cr & \int {\frac{{dy}}{{1 - 2y}}} = \int {dx} \cr & - \frac{1}{2}\log \left| {1 - 2y} \right| = x + C \cr & {\text{at }}x = 0,\,y = 0 \cr & - \frac{1}{2}\log \,1 = 0 + C \Rightarrow C = 0 \cr & 1 - 2y = {e^{ - 2x}}\,;\,y = \frac{{1 - {e^{ - 2x}}}}{2} \cr} $$