Differential Equations MCQ Questions & Answers in Calculus | Maths

$$y = a{e^{bx + c}} = a{e^c}.{e^{bx}} = k{e^{bx}}.$$
So, there are two arbitrary constants. Hence, the order of the differential equation will be 2.

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{\sqrt {1 - {y^2}} }}{y} \cr & \Rightarrow \frac{{ - 2y}}{{\sqrt {1 - {y^2}} }}dy + 2dx = 0 \cr & \Rightarrow 2\sqrt {1 - {y^2}} + 2x = 2c \cr & \Rightarrow \sqrt {1 - {y^2}} + x = c \cr & \Rightarrow {\left( {x - c} \right)^2} + {y^2} = 1 \cr} $$
which is a circle of fixed radius 1 and variable centre $$\left( {c,\,0} \right)$$   lying on $$x$$-axis.

Given differential equation,
$$\frac{{{d^2}y}}{{d{x^2}}} + x.\frac{{dy}}{{dx}} + \sin \,y + {x^2} = 0$$
The order of highest derivative $$ = 2$$  and degree $$ = 1.$$

$$\eqalign{ & \frac{{dy}}{{dx}} = \left| x \right| \cr & \frac{{dy}}{{dx}} = x{\text{ for }}x \geqslant 0\,;\frac{{dy}}{{dx}} = - x{\text{ for }}x < 0\,;\,\int {dy = } \int {x\,dx} \cr & y = \frac{{{x^2}}}{2} + {C_1}......\left( {{\text{i}}} \right) \cr & \int {dy = } - 1\,x\,dx \cr & y = - \frac{{{x^2}}}{2} + {C_1}......\left( {{\text{ii}}} \right) \cr & {\text{From }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & y = \frac{{x\left| x \right|}}{2} + C \cr} $$

$$\eqalign{ & \left( {x{y^3} - {x^2}} \right)dy - \left( {xy + {y^4}} \right)dx = 0 \cr & \Rightarrow {y^3}\left( {x\,dy - y\,dx} \right) - x\left( {x\,dy + y\,dx} \right) = 0 \cr & \Rightarrow {x^2}{y^3}\frac{{\left( {x\,dy - y\,dx} \right)}}{{{x^2}}} - x\left( {x\,dy + y\,dx} \right) = 0 \cr & \Rightarrow {x^2}{y^3}d\left( {\frac{y}{x}} \right) - xd\left( {xy} \right) = 0 \cr} $$
Dividing by $${x^3}{y^2},$$  we get $$\frac{y}{x}d\left( {\frac{y}{x}} \right) - \frac{{d\left( {xy} \right)}}{{{x^2}{y^2}}} = 0$$
Now, integrating $$\frac{1}{2}{\left( {\frac{y}{x}} \right)^2} + \frac{1}{{xy}} = c$$
It passes through the point $$\left( {4,\, - 2} \right)$$
$$\eqalign{ & \Rightarrow \frac{1}{8} - \frac{1}{8} = c \Rightarrow c = 0 \cr & \therefore \,{y^3} = - 2x \cr} $$

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

$$\eqalign{ & \because \,\frac{{dy}}{{dx}} = \sqrt {1 - {x^2} - {y^2} + {x^2}{y^2}} \cr & \frac{{dy}}{{dx}} = \sqrt {\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)} \,; \cr & \Rightarrow \frac{{dy}}{{\sqrt {1 - {y^2}} }} = \sqrt {1 - {x^2}} .dx \cr & \Rightarrow \int {\frac{{dy}}{{\sqrt {1 - {y^2}} }}} = \int {\sqrt {1 - {x^2}} .dx} \,\,\,\left[ {{\text{integrating }}\frac{b}{s}} \right] \cr & \Rightarrow {\sin ^{ - 1}}\left( {\frac{y}{1}} \right) = \frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{x}{1}} \right) + c \cr & \Rightarrow 2\,{\sin ^{ - 1}}y = x\sqrt {1 - {x^2}} + {\sin ^{ - 1}}x + c \cr} $$

$$\eqalign{ & {y^3} - x = cy \cr & \Rightarrow 3{y^2}\frac{{dy}}{{dx}} - 1 = c\frac{{dy}}{{dx}} \cr & \Rightarrow \frac{{dy}}{{dx}}\left( {3{y^2} - c} \right) = 1 \cr & \Rightarrow \frac{{dy}}{{dx}}\left( {3{y^2} - \frac{{{y^3} - x}}{y}} \right) = 1 \cr & \Rightarrow \frac{{dy}}{{dx}}\left( {\frac{{3{y^3} - {y^3} + x}}{y}} \right) = 1 \cr & \Rightarrow \frac{{dy}}{{dx}} = \frac{y}{{x + 2{y^3}}} \cr} $$

Statement 1 : Differential equation is not a polynomial equation in its derivatives. So, its degree is not defined.
Statement 2 : The highest order derivative in the given polynomial is 2.

The given equation is reduced to
$$\eqalign{ & x = {e^{xy\left( {\frac{{dy}}{{dx}}} \right)}} \cr & \Rightarrow \ell n\,x = xy\frac{{dy}}{{dx}} \cr & \Rightarrow \int {y\,dy} = \int {\frac{1}{x}} \ell n\,x\,dx \cr & \Rightarrow \frac{{{y^2}}}{2} = \frac{{{{\left( {\ell n\,x} \right)}^2}}}{2} + c \cr & \Rightarrow y = \pm \sqrt {{{\left( {\ell n\,x} \right)}^2}} + c \cr & \Rightarrow y = \pm \ell n\,x + c \cr} $$