Differential Equations MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & y = a + b{e^{5x}} + c{e^{ - 7x}}......\left( {\text{i}} \right) \cr & \therefore \,{y_1} = 0 + 5\,b{e^{5x}} - 7\,c{e^{ - 7x}} \cr} $$
Dividing by $$y{e^{5x}},$$  we get : $${e^{ - 5x}}{y_1} = 5b - 7\,c{e^{ - 12x}}$$
Again differentiating both sides w.r.t. $$x,$$ we get $${e^{ - 5x}}.{y_2} + {y_1}\left( { - 5} \right){e^{ - 5x}} = 0 + 84\,c{e^{ - 12x}}$$
Dividing by $${e^{ - 12x}}$$  we get : $${e^{7x}}\left( {{y_2} - 5{y_1}} \right) = 84\,c$$
Differentiating both sides w.r.t. $$x,$$ we get
$$\eqalign{ & {e^{7x}}\left( {{y_3} - 5{y_2}} \right) + \left( {{y_2} - 5{y_1}} \right).7\,{e^{7x}} = 0 \cr & \Rightarrow {y_3} + 2{y_2} - 35{y_1} = 0 \cr} $$

Divide the equation by $$\sqrt y ,$$ we get $${y^{ - \frac{1}{2}}}\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}{y^{\frac{1}{2}}} = x$$
$$\eqalign{ & {\text{Put }}{y^{\frac{1}{2}}} = z \cr & \Rightarrow \frac{1}{2}{y^{ - \frac{1}{2}}}\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}} \cr & \Rightarrow 2\frac{{dz}}{{dx}} + \frac{x}{{1 - {x^2}}}z = x \cr & \Rightarrow \frac{{dz}}{{dx}} + \left( {\frac{1}{2}\frac{x}{{1 - {x^2}}}} \right)z = \frac{x}{2} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{1}{2}} \left[ {\frac{x}{{1 - {x^2}}}} \right]dx}} = {e^{ - \frac{1}{4}\log \left( {1 - {x^2}} \right)}} = {\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} \cr & {\text{The solution is}} \cr & z{\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} = \int {\frac{x}{2}{{\left( {1 - x} \right)}^{ - \frac{1}{4}}}dx + c} \cr & \Rightarrow z{\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} = \frac{1}{2}\int {{{\left( {1 - {x^2}} \right)}^{ - \frac{1}{4}}}x\,dx + c} \cr & \Rightarrow \sqrt y {\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} = \frac{1}{2}\left( { - \frac{1}{2}} \right)\frac{{{{\left( {1 - {x^2}} \right)}^{\frac{3}{4}}}}}{{\frac{3}{4}}} + c \cr & \Rightarrow \sqrt y {\left( {1 - {x^2}} \right)^{ - \frac{1}{4}}} = - \frac{1}{3}{\left( {1 - {x^2}} \right)^{\frac{3}{4}}} + c \cr & \Rightarrow \sqrt y + \frac{1}{3}\left( {1 - {x^2}} \right) = c{\left( {1 - {x^2}} \right)^{\frac{1}{4}}} \cr} $$

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

General equation of circles passing through origin and having their centres on the $$x$$-axis is
$${x^2} + {y^2} + 2gx = 0\,.....({\text{i}})$$
On differentiating w.r.t $$x,$$ we get
$$2x + 2y.\frac{{dy}}{{dx}} + 2g = 0\,\,\,\, \Rightarrow g = - \left( {x + y\frac{{dy}}{{dx}}} \right)$$
$$\therefore $$ equation (i) be
$$\eqalign{ & {x^2} + {y^2} + 2\left\{ { - \left( {x + y\frac{{dy}}{{dx}}} \right)} \right\}x = 0 \cr & \Rightarrow {x^2} + {y^2} - 2{x^2} - 2x\frac{{dy}}{{dx}}.y = 0 \cr & \Rightarrow {y^2} = {x^2} + 2xy\frac{{dy}}{{dx}} \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} - \frac{t}{{1 + t}}y = \frac{1}{{1 + t}}\,\,\,{\text{which is in the linear form}}{\text{.}} \cr & {\text{IF }} = {e^{\int { - \,\frac{t}{{1 + t}}dt} }} = {e^{ - t + \log \,\left( {1 + t} \right)}} = \left( {1 + t} \right){e^{ - t}} \cr & \therefore y\left( {1 + t} \right){e^{ - t}} = \int {\frac{1}{{1 + t}}.\left( {1 + t} \right){e^{ - t}}dt} = \int {{e^{ - t}}dt} = - {e^{ - t}} + c \cr & {\text{When }}t = 0,\,y = - 1\,\,\,\,{\text{So, }} - 1 = - 1 + c\,\,\,\,\,\,{\text{Hence,}}\,\,c = 0 \cr & \therefore y\left( {1 + t} \right){e^{ - t}} = - {e^{ - t}} \cr & {\text{So, when }}t = 1,\,y.2.{e^{ - 1}} = - {e^{ - 1}} \cr & \therefore \,y = - \frac{1}{2} \cr} $$

[Note : derivative of a function is sometimes called as the marginal value, specially in economics]
Given $$\frac{{dc}}{{dx}} = 2 + 0.15x$$
Integrating we get
$$\eqalign{ & c\left( x \right) = 2x + 0.15\frac{{{x^2}}}{2} + D \cr & {\text{when }}x = 0,\,c\left( x \right) = 100 \Rightarrow D = 100 \cr} $$
$$\therefore $$  the required equation is $$c\left( x \right) = 0.075{x^2} + 2x + 100$$

$$\eqalign{ & \left( {1 + \log \,x} \right)\frac{{dx}}{{dy}} - x\,\log \,x = {e^y} \cr & {\text{putting }}x\,\log \,x = t \Rightarrow \left( {1 + \log \,x} \right)dx = dt \cr & \therefore \,\frac{{dt}}{{dy}} - t = {e^y}\,\,\,\,\,{\text{Now, I}}{\text{.F}}{\text{.}} = {e^{\int { - 1\,dy} }} = {e^{ - y}} \cr & \Rightarrow t{e^{ - y}} = \int {{e^{ - y}}{e^y}dy + C} \cr & \Rightarrow t = C{e^y} + y{e^y} \cr & \Rightarrow x\,\log \,x = \left( {C + y} \right){e^y}, \cr & {\text{Since, }}y\left( 1 \right) = 0,\,{\text{then}} \cr & 0 = \left( {C + 0} \right)1 \Rightarrow C = 0 \cr & \therefore \,y{e^y} = x\,\log \,x \Rightarrow {x^x} = {e^{y{e^y}}} \cr} $$

$$\eqalign{ & \left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - \,1}}y}}} \right)\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow \frac{{dx}}{{dy}} + \frac{x}{{\left( {1 + {y^2}} \right)}} = \frac{{{e^{\,{{\tan }^{ - \,1}}y}}}}{{\left( {1 + {y^2}} \right)}} \cr & I.F. = {e^{\int {\frac{1}{{\left( {1 + {y^2}} \right)}}\,dy} }} = {e^{\,{{\tan }^{ - \,1}}y}} \cr & x\left( {{e^{\,{{\tan }^{ - \,1}}y}}} \right) = \int {\frac{{{e^{{{\tan }^{ - \,1}}y}}}}{{1 + {y^2}}}{e^{{{\tan }^{ - \,1}}y\,\,dy}}} \cr & x\left( {{e^{\,{{\tan }^{ - \,1}}y}}} \right) = \frac{{{e^{2\,{{\tan }^{ - \,1}}y\,}}}}{2} + C \cr & \therefore \,2x{e^{{{\tan }^{ - \,1}}y\,}} = {e^{2\,{{\tan }^{ - \,1}}y\,}} + k \cr} $$

$$\eqalign{ & {x^2} + {y^2} - 2ay = 0\,.....\left( 1 \right) \cr & {\text{Differentiate,}} \cr & 2x + 2y\frac{{dy}}{{dx}} - 2a\frac{{dy}}{{dx}} = 0\,\, \Rightarrow a = \frac{{x + yy'}}{{y'}} \cr & {\text{Put in }}\left( 1 \right){\text{,}}\,\,{x^2} + {y^2} - 2\left( {\frac{{x + yy'}}{{y'}}} \right)y = 0 \cr & \Rightarrow \left( {{x^2} + {y^2}} \right)y' - 2xy - 2{y^2}y' = 0 \cr & \Rightarrow \left( {{x^2} - {y^2}} \right)y' = 2xy \cr} $$

Given curve is $${y^2} = 4a\left( {x - a} \right)......\left( {\text{i}} \right)$$
On differentiating w.r.t. $$x,$$ we get
$$2yy' = 4a \Rightarrow a = \frac{{yy'}}{2}$$
On putting the value of $$a$$ in Equation $$\left( {\text{i}} \right),$$ we get
$$\eqalign{ & {y^2} = 4\left( {\frac{{yy'}}{2}} \right)\left( {x - \frac{{yy'}}{2}} \right) \cr & \Rightarrow {y^2} = yy'\left( {2x - yy'} \right) \cr & \Rightarrow yy'\left( {yy' - 2x} \right) + {y^2} = 0 \cr} $$