Differential Equations MCQ Questions & Answers in Calculus | Maths

The given equation can be written as $$\frac{y}{x}\frac{{dy}}{{dx}} = \left\{ {\frac{{{y^2}}}{{{x^2}}} + \frac{{f\left( {\frac{{{y^2}}}{{{x^2}}}} \right)}}{{f'\left( {\frac{{{y^2}}}{{{x^2}}}} \right)}}} \right\}$$
The above equation is a homogeneous equation.
Putting $$y = vx,$$   we get
$$\eqalign{ & v\left[ {v + x\frac{{dv}}{{dx}}} \right] = {v^2} + \frac{{f\left( {{v^2}} \right)}}{{f'\left( {{v^2}} \right)}} \cr & {\text{or }}vx\frac{{dv}}{{dx}} = \frac{{f\left( {{v^2}} \right)}}{{f'\left( {{v^2}} \right)}}\,\,\,\,\left( {{\text{variable separable}}} \right) \cr & {\text{or }}\frac{{2vf'\left( {{v^2}} \right)}}{{f\left( {{v^2}} \right)}}dv = 2\frac{{dx}}{x} \cr} $$
Now, integrating both sides, we get
$$\eqalign{ & \log \,f\left( {{v^2}} \right) = \log \,{x^2} + \log \,c\,\,\,\left[ {\log \,c = {\text{ constant}}} \right] \cr & {\text{or }}\log \,f\left( {{v^2}} \right) = \log \,c\,{x^2}\, \cr & {\text{or }}\,f\left( {{v^2}} \right) = c{x^2} \cr & {\text{or }}f\left( {\frac{{{y^2}}}{{{x^2}}}} \right) = c{x^2} \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} = y + 3 \Rightarrow \int {\frac{{dy}}{{y + 3}} = \int {dx} } \cr & \Rightarrow \ell n\left| {y + 3} \right| = x + c \cr & {\text{Since }}y\left( 0 \right) = 2\,\,\,\,\,\,\,\,\,\,\,\,\therefore \ell n\,5 = c \cr & \Rightarrow \ell n\left| {y + 3} \right| = x + \ell n\,5 \cr & {\text{When }}x = \ell n\,2,\,\,{\text{then }}\ell n\left| {y + 3} \right| = \ell n\,2 + \ell n\,5 \cr & \Rightarrow \ell n\left| {y + 3} \right| = \ell n\,10 \cr & \therefore y + 3 = \pm 10 \cr & \Rightarrow y = 7,\,\, - 13 \cr} $$

The given equation can be rewritten as
$$\eqalign{ & {x^2}\left[ {\frac{{x\,dy - y\,dx}}{{{x^2}}}} \right] = x\sqrt {1 - \frac{{{y^2}}}{{{x^2}}}} \cr & \Rightarrow {x^2}\frac{d}{{dx}}\left( {\frac{y}{x}} \right) = x\sqrt {1 - \frac{{{y^2}}}{{{x^2}}}} \cr & {\text{Put }}\frac{y}{x} = z,{\text{ we get}} \cr & \Rightarrow {x^2}\frac{{dz}}{{dx}} = x\sqrt {1 - {z^2}} \cr & \Rightarrow \frac{{dz}}{{\sqrt {1 - {z^2}} }} = \frac{{dx}}{x} \cr & {\text{Integrating, we get}} \cr & {\sin ^{ - 1}}\left( z \right) = \log \,x + c \cr & \Rightarrow {\sin ^{ - 1}}\left( {\frac{y}{x}} \right) = \log \,x + c \cr & {\text{Apply the boundary value}}\,y\left( 1 \right) = 0 \cr & \Rightarrow {\sin ^{ - 1}}\left( {\frac{0}{1}} \right) = \log \,1 + c \Rightarrow c = 0 \cr & \therefore \,{\sin ^{ - 1}}\left( {\frac{y}{x}} \right) = \log \,x + 0 \Rightarrow y = x\,\sin \left( {\log \,x} \right) \cr} $$

$$\eqalign{ & \left( {x + 1} \right)\frac{{dy}}{{dx}} + 1 = 2{e^{ - y}} \cr & \Rightarrow {e^y}\frac{{dy}}{{dx}} + \frac{{{e^y}}}{{x + 1}} = \frac{2}{{x + 1}} \cr & {\text{Put }}{e^y} = u \Rightarrow {e^y}\frac{{dy}}{{dx}} = \frac{{du}}{{dx}} \cr & \therefore \,\frac{{du}}{{dx}} + \frac{u}{{x + 1}} = \frac{2}{{x + 1}} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{1}{{x + 1}}dx} }} = {e^{\log \left( {x + 1} \right)}} = x + 1 \cr & \therefore {\text{ Solution is }} \cr & u\left( {x + 1} \right) = \int {\frac{2}{{\left( {x + 1} \right)}}\left( {x + 1} \right)dx + C} \cr & \Rightarrow {e^y}\left( {x + 1} \right) = 2x + C \cr} $$

The given equation is
$$\eqalign{ & \frac{{dy}}{{dx}} - \frac{y}{{x + 1}} = {e^{3x}}\left( {x + 1} \right) \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int { - \frac{1}{{x + 1}}dx} }} = {e^{ - \log \left( {x + 1} \right)}} = \frac{1}{{x + 1}} \cr & {\text{The solution is }} \cr & y\left( {\frac{1}{{x + 1}}} \right) = \int {{e^{3x}}\left( {x + 1} \right).\frac{1}{{x + 1}}dx + a} \cr & \Rightarrow \frac{y}{{x + 1}} = \int {{e^{3x}}dx + a} \cr & \Rightarrow \frac{y}{{x + 1}} = \frac{{{e^{3x}}}}{3} + a \cr & \Rightarrow \frac{{3y}}{{x + 1}} = {e^{3x}} + c,\,c = 3a \cr} $$

Given differential equation is
$$\eqalign{ & \frac{{dp\left( t \right)}}{{dt}} = 0.5\,p\left( t \right) - 450 \cr & \Rightarrow \frac{{dp\left( t \right)}}{{dt}} = \frac{1}{2}\,p\left( t \right) - 450 \cr & \Rightarrow \frac{{dp\left( t \right)}}{{dt}} = \frac{{p\left( t \right) - 900}}{2} \cr & \Rightarrow 2\frac{{dp\left( t \right)}}{{dt}} = - \left[ {900 - \,p\left( t \right)} \right] \cr & \Rightarrow 2\frac{{dp\left( t \right)}}{{900 - \,p\left( t \right)}} = - dt \cr} $$
Integrate both sides, we get
$$\eqalign{ & - 2\int {\frac{{dp\left( t \right)}}{{900 - \,p\left( t \right)}}} = \int {dt} \cr & {\text{Let }}900 - \,p\left( t \right) = u \cr & \Rightarrow - dp\left( t \right) = du \cr & \therefore {\text{We have, }}2\int {\frac{{du}}{u} = \int {dt} } \,\,\,\, \Rightarrow 2\,\ln \,u = t + c \cr & \Rightarrow 2\,\ln \left[ {900 - \,p\left( t \right)} \right] = t + c \cr & {\text{when }}t = 0,\,\,\,\,p\left( 0 \right) = 850 \cr & 2\,\ln \,\left( {50} \right) = c \cr & \Rightarrow 2\left[ {\ln \left( {\frac{{900 - \,p\left( t \right)}}{{50}}} \right)} \right] = t \cr & \Rightarrow 900 - \,p\left( t \right) = 50{e^{\frac{t}{2}}} \cr & \Rightarrow p\left( t \right) = 900 - \,50{e^{\frac{t}{2}}} \cr & {\text{Let }}p\left( {{t_1}} \right) = 0 \cr & 0 = 900 - - \,50{e^{\frac{{{t_1}}}{2}}} \cr & \therefore {t_1} = \,2\,\ln \,18 \cr} $$

$$\eqalign{ & \frac{{df\left( x \right)}}{{dx}} = \frac{{{e^{\sin \,x}}}}{x} \cr & \Rightarrow \frac{{df\left( {{x^3}} \right)}}{{d\left( {{x^3}} \right)}} = \frac{{{e^{\sin \,{x^3}}}}}{{{x^3}}} \cr & \Rightarrow \frac{{df\left( {{x^3}} \right)}}{{d\left( {{x^3}} \right)}}.\frac{{d\left( {{x^3}} \right)}}{{dx}} = \frac{{{e^{\sin \,{x^3}}}}}{{{x^3}}}.3{x^2} \cr & \Rightarrow \frac{{df\left( {{x^3}} \right)}}{{dx}} = \frac{{3{e^{\sin \,{x^3}}}}}{x} \cr & \Rightarrow \int_1^4 {\frac{{3{e^{\sin \,{x^3}}}}}{x}dx} = \left[ {f\left( {{x^3}} \right)} \right]_{x = 1}^4 = f\left( {64} \right) - f\left( 1 \right) \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{e^x}\left( {{{\sin }^2}x + \sin \,2x} \right)}}{{y\left( {2\,\log \,y + 1} \right)}} \cr & \Rightarrow \int {\left( {2y\,\log \,y + y} \right)dy} = \int {{e^x}\left( {{{\sin }^2}x + \sin \,2x} \right)dx} \cr} $$
On integrating by parts, we get $${y^2}\left( {\log \,y} \right) = {e^x}{\sin ^2}x + c.$$

$$\eqalign{ & \sin \,x\frac{{dy}}{{dx}} + y\,\cos \,x = 1 \cr & \frac{{dy}}{{dx}} + y\,\cot \,x = {\text{cosec}}\,x \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\cot \,x\,dx} }} = {e^{\ln \left( {\sin \,x} \right)}} = \sin \,x \cr & y\,\sin \,x = \int {{\text{cosec}}\,x.\sin \,x\,dx = x + C} \cr & {\text{If }}x = 0,\,y{\text{ is finite}} \cr & \therefore \,C = 0 \cr & y = x\left( {{\text{cosec}}\,x} \right) = \frac{x}{{\sin \,x}} \cr & {\text{Now, }}l < \frac{{{\pi ^2}}}{4}{\text{ and }}l > \frac{\pi }{2} \cr & {\text{Hence, }}\frac{\pi }{2} < l < \frac{{{\pi ^2}}}{4} \cr} $$

$$\frac{{dy}}{{dx}} + \frac{2}{x}y = x$$     and $$y\left( a \right) = 1$$   (Given)
Since, the above differential equation is the linear differential equation, then
$${\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{2}{x}\,dx} }} = {x^2}$$
Now, the solution of the linear differential equation
$$\eqalign{ & y \times {x^2} = \int {{x^3}dx\,\,} \Rightarrow y{x^2} = \frac{{{x^4}}}{4} + C \cr & \because y\left( 1 \right) = 1 \cr & \therefore 1 \times 1 = \frac{1}{4} + C\,\, \Rightarrow C = \frac{3}{4} \cr} $$
$$\therefore $$ solution becomes
$$y = \frac{{{x^2}}}{4} + \frac{3}{{4{x^2}}}$$