Differential Equations MCQ Questions & Answers in Calculus | Maths

Given D.E. can be written as
$$\eqalign{ & \frac{{dy}}{{dx}} - \frac{x}{{1 - {x^2}}}y = \frac{{{x^4} + 2x}}{{\sqrt {1 - {x^2}} }} \cr & {\text{If }}{e^{\int {\frac{{ - \,x}}{{1\, - \,{x^2}}}dx} }} = {e^{\frac{{ + \,1}}{2}\,\log \,\left( {1\, - \,{x^2}} \right)}} = \sqrt {1 - {x^2}} \cr} $$
$$\therefore $$ Solution is given by
$$\eqalign{ & y\sqrt {1 - {x^2}} = \int {\sqrt {1 - {x^2}} } .\frac{{{x^4} + 2x}}{{\sqrt {1 - {x^2}} }}dx \cr & y\sqrt {1 - {x^2}} = \frac{{{x^5}}}{5} + {x^2} + c \cr & f\left( 0 \right) = 0\,\,\, \Rightarrow {\text{At }}x = 0,\,\,y = 0 \cr & \therefore c = 0 \cr & \therefore y\sqrt {1 - {x^2}} = \frac{{{x^5}}}{5} + {x^2} \cr & {\text{or,}}\,y = f\left( x \right) = \frac{{\frac{{{x^5}}}{5} + {x^2}}}{{\sqrt {1 - {x^2}} }} \cr & \therefore \,I = \int\limits_{ - \,\frac{{\sqrt 3 }}{2}}^{\frac{{\sqrt 3 }}{2}} {\frac{{\frac{{{x^5}}}{5} + {x^2}}}{{\sqrt {1 - {x^2}} }}dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 2\int\limits_0^{\frac{{\sqrt 3 }}{2}} {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}dx} \,\,\left( {\because \frac{{{x^5}}}{{\sqrt {1 - {x^2}} }}{\text{ is odd}}} \right) \cr & {\text{Put}}\,\,x = \sin \theta \,\,\,\,\,\, \Rightarrow dx = \cos \theta \,\,d\theta \, \cr & \therefore \,\,I = 2\int\limits_0^{\frac{\pi }{3}} {{{\sin }^2}\theta d\theta } \cr & = \int\limits_0^{\frac{\pi }{3}} {\left( {1 - \cos \,2\theta } \right)d\theta } \, \cr & = \left( {\theta - \frac{{\sin \,2\theta }}{2}} \right)_0^{\frac{\pi }{3}} \cr & = \frac{\pi }{3} - \frac{{\sqrt 3 }}{4} \cr} $$

Given, differential equation as $$1 + 3{\left( {\frac{{dy}}{{dx}}} \right)^{\frac{2}{3}}} = 4\frac{{{d^3}y}}{{d{x^3}}}$$
On cubing the equation on both the sides,
$$\eqalign{ & {\left( {1 + 3{{\left( {\frac{{dy}}{{dx}}} \right)}^{\frac{2}{3}}}} \right)^3} = {\left( {4\frac{{{d^3}y}}{{d{x^3}}}} \right)^3} \cr & \Rightarrow {\left( {1 + 3\frac{{dy}}{{dx}}} \right)^2} = 64 \times {\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^3} \cr} $$
The order of the equation is the highest degree of differential in the equation.
Here it is 3....... (since $$\frac{{{d^3}y}}{{d{x^3}}}$$   exists in the equation)
∴ Order $$=3$$
The degree of the equation is the power of the highest order differential term in the equation.
Here, it is 3........ (since power of $$\frac{{{d^3}y}}{{d{x^3}}}$$   in the equation is 3)
∴ Degree $$=3$$

This is the form in which $$\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}}$$
The given equation can be rewritten as $$\frac{{dy}}{{dx}} = \frac{{1 - 3\left( {x + y} \right)}}{{1 + \left( {x + y} \right)}} = f\left( {x + y} \right)$$
Substitute $$x + y = z \Rightarrow 1 + \frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}$$
The equation then becomes
$$\eqalign{ & \frac{{dz}}{{dx}} - 1 = \frac{{1 - 3z}}{{1 + z}} \cr & \Rightarrow \frac{{dz}}{{dx}} = \frac{{1 - 3z + 1 + z}}{{1 + z}} \cr & \Rightarrow \frac{{dz}}{{dx}} = \frac{{2 - 2z}}{{1 + z}} \cr & \Rightarrow \frac{{1 + z}}{{2\left( {1 - z} \right)}}dz = dx \cr} $$
On integrating we get
$$\eqalign{ & \frac{1}{2}\int {\frac{{1 + z}}{{1 - z}}dz = } \int {dx + a} \cr & \Rightarrow \frac{1}{2}\int {\left[ {\frac{2}{{1 - z}} - 1} \right]dz = x + a} \cr & \Rightarrow - \ell n\left| {1 - z} \right| - \frac{1}{2}z = x + a \cr & \Rightarrow - \ell n\left| {1 - x - y} \right| - \frac{1}{2}\left( {x + y} \right) = x + a \cr & \Rightarrow - 2\ell n\left| {1 - x - y} \right| - 3x - y = 2a \cr & \Rightarrow 3x + y + 2\ell n\left| {1 - x - y} \right| = c{\text{ where }}c = - 2a \cr} $$

Degree of differential equation is always a positive integer.
$$\therefore \,\left( {2,\,\frac{3}{2}} \right)$$   can not be the feasible.

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