NCERT Solutions Class 12 Maths (Determinants) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-9 (Determinants)Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Q1. $\begin{array}{l} For each of the differential equations given below, indicate its order and degree (if \\ defined). \\ (i) \frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = \log x \\ (i i) {(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = \sin x \\ (i i i) \frac{d^{4} y}{d x^{4}} - \sin (\frac{d^{3} y}{d x^{3}}) = 0 \end{array}$

Answer. $\begin{array}{l} (i) The differential equation is given as: \\ \frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = \log x \\ \Rightarrow \frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y - \log x = 0 \\ The highest order derivative present in the differential equation is one. \\ The highest power raised to \frac{d^{2} y}{d x^{2}} is one. Hence, its degree is one. \end{array}$ $\begin{array}{l} (ii) The differential equation is given as: \\ {(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = \sin x \\ \Rightarrow {(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y - \sin x = 0 \\ The highest order derivative present in the differential equation is \frac{d y}{d x} . Thus, its order is one. \\ The highest power raised to \frac{d y}{d x} is three. Hence, its degree is three. \end{array}$ $\begin{array}{l} (iii) The differential equation is given as: \\ \frac{d^{4} y}{d x^{4}} - \sin (\frac{d^{3} y}{d x^{3}}) = 0 \\ The highest order derivative present in the differential equation is \frac{d^{4} y}{d x^{4}}, Thus, its order is four. \\ However, the given differential equation is not a polynomial equation. Hence, its order is not \\ defined. \end{array}$

Q2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. $\begin{array}{l} (i) y = a e^{x} + b e^{- x} + x^{2} : x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0 \\ (i i) y = e^{x} (a \cos x + b \sin x) : \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0 \end{array}$ $\begin{array}{ll} (i i i) y = x \sin 3 x & : \frac{d^{2} y}{d x^{2}} + 9 y - 6 \cos 3 x = 0 \\ (i v) x^{2} = 2 y^{2} \log y & : (x^{2} + y^{2}) \frac{d y}{d x} - x y = 0 \end{array}$

Answer. $\begin{array}{l} (i) y = a e^{x} + b e^{- x} + x^{2} \\ Differentiating both sides with respect to x, we get: \\ \frac{d y}{d x} = a \frac{d}{d x} (e^{x}) + b \frac{d}{d x} (e^{- x}) + \frac{d}{d x} (x^{2}) \\ \Rightarrow \frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x \\ Again, differentiating both sides with respect to x, we get: \\ \frac{d^{2} y}{d x^{2}} = a e^{x} + b e^{- x} + 2 \end{array}$ $\begin{array}{l} Now, on substituting the values of \frac{d y}{d x} and \frac{d^{2} y}{d x^{2}} in the differential equation, we get: \\ L.H.S. \\ x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 \end{array}$ $\begin{aligned} = x (a e^{x} + b e^{- x} + 2) + 2 (a e^{x} - b e^{- x} + 2 x) - x (a e^{x} + b e^{- x} + x^{2}) + x^{2} - 2 \\ = (a x e^{x} + b x e^{- x} + 2 x) + (2 a e^{x} - 2 b e^{- x} + 4 x) - (a x e^{x} + b x e^{- x} + x^{3}) + x^{2} - 2 \\ = 2 a e^{x} - 2 b e^{- x} + x^{2} + 6 x - 2 \\ \neq 0 \end{aligned}$ ⇒ L.H.S. ≠ R.H.S. Hence, the given function is not a solution of the corresponding differential equation. $\begin{array}{l} (i i) y = e^{x} (a \cos x + b \sin x) = a e^{x} \cos x + b e^{x} \sin x \\ Differentiating both sides with respect to x, we get: \\ \frac{d y}{d x} = a \cdot \frac{d}{d x} (e^{x} \cos x) + b \cdot \frac{d}{d x} (e^{x} \sin x) \\ \Rightarrow \frac{d y}{d x} = a (e^{x} \cos x - e^{x} \sin x) + b \cdot (e^{x} \sin x + e^{x} \cos x) \\ \Rightarrow \frac{d y}{d x} = (a + b) e^{x} \cos x + (b - a) e^{x} \sin x \\ A g a i n, d i f f e r e n t i a t i n g b o t h s i d e s w i t h r e s p e c t t o x, w e g e t : \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = (a + b) \cdot \frac{d}{d x} (e^{x} \cos x) + (b - a) \frac{d}{d x} (e^{x} \sin x) \end{array}$ $\begin{array}{l} \Rightarrow \frac{d^{2} y}{d x^{2}} = (a + b) \cdot [e^{x} \cos x - e^{x} \sin x] + (b - a) [e^{x} \sin x + e^{x} \cos x] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} [(a + b) (\cos x - \sin x) + (b - a) (\sin x + \cos x)] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} [a \cos x - a \sin x + b \cos x - b \sin x + b \sin x + b \cos x - a \sin x - a \cos x] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = [2 e^{x} (b \cos x - a \sin x)] \end{array}$ $\begin{array}{l} Now, on substituting the values of \frac{d^{2} y}{d x^{2}} and \frac{d y}{d x} in the L.H.S. of the given differential \\ equation, we get: \\ \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + 2 y \end{array}$ $\begin{aligned} = 2 e^{x} (b \cos x - a \sin x) - 2 e^{x} [(a + b) \cos x + (b - a) \sin x] + 2 e^{x} (a \cos x + b \sin x) \\ = e^{x} [(2 b \sin x - 2 a \sin x) - (2 a \cos x + 2 b \cos x) \\ = e^{x} [(2 b \sin x - 2 a \sin x) + (2 a \cos x + 2 b \sin x)] \\ = e^{x} [(2 b - 2 a - 2 b + 2 a) \cos x] + e^{x} [(- 2 a - 2 b + 2 a + 2 b) \sin x] \\ = 0 \end{aligned}$ Hence, the given function is a solution of the corresponding differential equation. $\begin{array}{l} (i i i) y = x \sin 3 x \\ Differentiating both sides with respect to x, we get: \\ \frac{d y}{d x} = \frac{d}{d x} (x \sin 3 x) = \sin 3 x + x \cdot \cos 3 x \cdot 3 \\ \Rightarrow \frac{d y}{d x} = \sin 3 x + 3 x \cos 3 x \\ Again, differentiating both sides with respect to x, we get: \\ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\sin 3 x) + 3 \frac{d}{d x} (x \cos 3 x) \end{array}$ $\begin{array}{l} \Rightarrow \frac{d^{2} y}{d x^{2}} = 3 \cos 3 x + 3 [\cos 3 x + x (- \sin 3 x) \cdot 3] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = 6 \cos 3 x - 9 x \sin 3 x \\ Substituting the value of \frac{d^{2} y}{d x^{2}} in the L.H.S. of the given differential equation, we get: \\ \frac{d^{2} y}{d x^{2}} + 9 y - 6 \cos 3 x \end{array}$ $\begin{array}{l} = (6 \cdot \cos 3 x - 9 x \sin 3 x) + 9 x \sin 3 x - 6 \cos 3 x \\ = 0 \\ Hence, the given function is a solution of the corresponding differential equation. \end{array}$ $\begin{array}{l} (iv) x^{2} = 2 y^{2} \log y \\ Differentiating both sides with respect to x, we get: \\ 2 x = 2 \cdot \frac{d}{d x} = [y^{2} \log y] \\ \Rightarrow x = [2 y \cdot \log y \cdot \frac{d y}{d x} + y^{2} \cdot \frac{1}{y} \cdot \frac{d y}{d x}] \\ \Rightarrow x = \frac{d y}{d x} (2 y \log y + y) \\ \Rightarrow \frac{d y}{d x} = \frac{x}{y (1 + 2 \log y)} \end{array}$ $\begin{array}{l} Substituting the value of \frac{d y}{d x} in the L.H.S. of the given differential equation, we get: \\ (x^{2} + y^{2}) \frac{d y}{d x} - x y \end{array}$ $\begin{array}{l} = (2 y^{2} \log y + y^{2}) \cdot \frac{x}{y (1 + 2 \log y)} - x y \\ = y^{2} (1 + 2 \log y) \cdot \frac{x}{y (1 + 2 \log y)} - x y \\ = x y - x y \\ = 0 \\ Hence, the given function is a solution of the corresponding differential equation. \end{array}$

Q3. $\begin{array}{l} Form the differential equation representing the family of curves given by \\ (x - a)^{2} + 2 y^{2} = a^{2} where a is an arbitrary constant. \end{array}$

Answer. $\begin{array}{l} (x - a)^{2} + 2 y^{2} = a^{2} \\ \Rightarrow x^{2} + a^{2} - 2 a x + 2 y^{2} = a^{2} \\ \Rightarrow 2 y^{2} = 2 a x - x^{2} . . . (i) \\ Differentiating with respect to x, we get: \\ 2 y \frac{d y}{d x} = \frac{2 a - 2 x}{2} \\ \Rightarrow \frac{d y}{d x} = \frac{2 a x - 2 x^{2}}{2 x y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 a x - 2 x^{2}}{4 x y} . . . (i i) \end{array}$ $\begin{array}{l} From equation (i), we get: \\ 2 a x = 2 y^{2} + x^{2} \\ On substituting this value in equation (i i), we get: \\ \frac{d y}{d x} = \frac{2 y^{2} + x^{2} - 2 x^{2}}{4 x y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 y^{2} - x^{2}}{4 x y} \end{array}$ Hence, the differential equation of the family of curves is given as $\frac{d y}{d x} = \frac{2 y^{2} - x^{2}}{4 x y}$

Q4. $\begin{array}{l} Prove that x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2} is the general solution of differential equation \\ (x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y \end{array}$ where c is a parameter.

Answer. $\begin{array}{l} (x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y \\ \Rightarrow \frac{d y}{d x} = \frac{x^{3} - 3 x y^{2}}{y^{3} - 3 x^{2} y} . . . (i) \\ This is a homogeneous equation. To simplify it, we need to make the substitution as: y=vx \\ \Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (v x) \\ \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \end{array}$ $\begin{array}{l} Substituting the values of y and \frac{d v}{d x} in equation (i), we get: \\ v + x \frac{d v}{d x} = \frac{x^{3} - 3 x (v x)^{2}}{(v x)^{3} - 3 x^{2} (v x)} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} - v \end{array}$ $\begin{array}{l} \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2} - v (v^{3} - 3 v)}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - v^{4}}{v^{3} - 3 v} \\ \Rightarrow (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \frac{d x}{x} \end{array}$ $\begin{array}{l} Integrating both sides, we get: \\ \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \log x + \log C^{'} (i i) \\ Now, \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \int \frac{v^{3} d v}{1 - v^{4}} - 3 \int \frac{v d v}{1 - v^{4}} \\ \Rightarrow \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = I_{1} - 3 I_{2}, where I_{1} = \int \frac{v^{3} d v}{1 - v^{4}} and I_{2} = \int \frac{v d v}{1 - v^{4}} . . . (i i i) \end{array}$ $\begin{array}{l} Let 1 - v^{4} = t \\ ∴ \frac{d}{d v} (1 - v^{4}) = \frac{d t}{d v} \\ \Rightarrow - 4 v^{3} = \frac{d t}{d v} \\ \Rightarrow v^{3} d v = - \frac{d t}{4} \\ Now, I_{1} = \int \frac{- d t}{4 t} = - \frac{1}{4} \log t = - \frac{1}{4} \log (1 - v^{4}) \\ And, I_{2} = \int \frac{v d v}{d v} = \int \frac{v d v}{1 - {(v^{2})}^{2}} \\ ∴ \frac{d}{d v} (v^{2}) = \frac{d p}{d v} \end{array}$ $\begin{array}{l} \Rightarrow 2 v = \frac{d p}{d v} \\ \Rightarrow v d v = \frac{d p}{2} \\ \Rightarrow I_{2} = \frac{1}{2} \int \frac{d p}{1 - p^{2}} = \frac{1}{2 \times 2} \log | \frac{1 + p}{1 - p} | = \frac{1}{4} \log | \frac{1 + v^{2}}{1 - v^{2}} | \\ Substituting the values of I_{1} and I_{2} in equation (i i i), we get: \\ \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = - \frac{1}{4} \log (1 - v^{4}) - \frac{3}{4} \log \frac{| - v^{2}}{1 + v^{2}} | \\ Therefore, equation (i i) becomes: \end{array}$ $\begin{array}{l} \frac{1}{4} \log (1 - v^{4}) - \frac{3}{4} \log \frac{| 1 + v^{2} |}{1 - v^{2}} = \log x + \log C^{'} \\ \Rightarrow - \frac{1}{4} \log [(1 - v^{4}) {(\frac{1 + v^{2}}{1 - v^{2}})}^{3}] = \log C^{'} x \\ \Rightarrow \frac{{(1 + v^{2})}^{4}}{{(1 - v^{2})}^{2}} = {(C^{'} x)}^{- 4} \end{array}$ $\begin{array}{l} \Rightarrow \frac{{(1 + \frac{y^{2}}{x^{2}})}^{4}}{{(1 - \frac{y^{2}}{x^{2}})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow \frac{{(x^{2} + y^{2})}^{4}}{x^{4} {(x^{2} - y^{2})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow (x^{2} - y^{2}) = C^{' 4} {(x^{2} + y^{2})}^{2} \\ \Rightarrow x^{2} - y^{2} = C^{' 2} {(x^{2} + y^{2})}^{2} \\ \Rightarrow x^{2} - y^{2} = C {(x^{2} + y^{2})}^{2}, where C = C^{' 2} \\ Hence, the given result is proved. \end{array}$

Q5. Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Answer. $\begin{array}{l} The equation of a circle in the first quadrant with centre (a, a) and radius (a) which \\ touches the coordinate axes is: \\ (x - a)^{2} + (y - a)^{2} = a^{2} . . . (i) \end{array}$ Solutions Class 12 Maths Chapter-9 (Determinants)Miscellaneous Exercise $\begin{array}{l} Differentiating equation (i) with respect to x, we get: \\ 2 (x - a) + 2 (y - a) \frac{d y}{d x} = 0 \\ \Rightarrow (x - a) + (y - a) y^{'} = 0 \\ \Rightarrow x - a + y y^{'} - a y^{'} = 0 \\ \Rightarrow x + y y^{'} - a (1 + y^{'}) = 0 \\ \Rightarrow a = \frac{x + y y^{'}}{1 + y^{'}} \end{array}$ $\begin{array}{l} Substituting the value of a in equation (i), we get: \\ {[x - (\frac{x + y y^{'}}{1 + y^{'}})]}^{2} + {[y - (\frac{x + y y^{'}}{1 + y^{'}})]}^{2} = {(\frac{x + y y^{'}}{1 + y^{'}})}^{2} \\ \Rightarrow {[\frac{(x - y) y^{'}}{(1 + y^{'})}]}^{2} + {[\frac{y - x}{1 + y^{'}}]}^{2} = {[\frac{x + y y^{'}}{1 + y^{'}}]}^{2} \\ \Rightarrow (x - y)^{2} \cdot y^{2} + (x - y)^{2} = {(x + y y^{'})}^{2} \\ \Rightarrow (x - y)^{2} [1 + {(y^{'})}^{2}] = {(x + y y^{'})}^{2} \\ H e n c e, t h e r e q u i r e d d i f f e r e n t i a l e q u a t i o n o f t h e f a m i l y o f c i r c l e s a r e \\ (x - y)^{2} [1 + {(y^{'})}^{2}] = {(x + y y^{'})}^{2} \end{array}$

Q6. Find the general solution of the differential equation $d x^{\frac{d y}{d x}} + \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} = 0$

Answer. $\begin{array}{l} \frac{d y}{d x} + \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} = 0 \\ \Rightarrow \frac{d y}{d x} = - \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}} \\ \Rightarrow \frac{d y}{\sqrt{1 - y^{2}}} = \frac{- d x}{\sqrt{1 - x^{2}}} \\ Integrating both sides, we get: \\ \sin^{- 1} y = - \sin^{- 1} x + C \\ \Rightarrow \sin^{- 1} x + \sin^{- 1} y = C \end{array}$

Q7. $\begin{array}{l} Show that the general solution of the differential equation \frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0 \\ (x + y + 1) = A (1 - x - y - 2 x y), s, where A is parameter \end{array}$

Answer. $\begin{array}{l} \frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0 \\ \Rightarrow \frac{d y}{d x} = - \frac{(y^{2} + y + 1)}{x^{2} + x + 1} \\ \Rightarrow \frac{d y}{y^{2} + y + 1} = \frac{- d x}{x^{2} + x + 1} \\ \Rightarrow \frac{d y}{y^{2} + y + 1} + \frac{d x}{x^{2} + x + 1} = 0 \\ Integrating both sides, we get: \end{array}$ $\begin{array}{l} \int \frac{d y}{y^{2} + y + 1} + \int \frac{d x}{x^{2} + x + 1} = C \\ \Rightarrow \int \frac{d y}{{(y + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} + \int \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = C \\ \Rightarrow \frac{2}{\sqrt{3}} \tan^{- 1} [\frac{y + \frac{1}{2}}{2}] + \frac{2}{\sqrt{3}} \tan^{- 1} [\frac{x + \frac{1}{2}}{2}] = C \\ \Rightarrow \tan^{- 1} [\frac{2 y + 1}{\sqrt{3}}] + \tan^{- 1} [\frac{2 x + 1}{\sqrt{3}}] = \frac{\sqrt{3} C}{2} \end{array}$ $\begin{array}{l} \Rightarrow \tan^{- 1} [\frac{\frac{2 y + 1}{\sqrt{3}} + \frac{2 x + 1}{\sqrt{3}}}{1 - \frac{(2 y + 1)}{\sqrt{3}} \cdot \frac{(2 x + 1)}{\sqrt{3}}}] = \frac{\sqrt{3} C}{2} \\ \Rightarrow \tan^{- 1} [\frac{\frac{2 x + 2 y + 2}{\sqrt{3}}}{1 - (\frac{4 x y + 2 x + 2 y + 1}{3})}] = \frac{\sqrt{3} C}{2} \end{array}$ $\begin{array}{l} \Rightarrow \tan^{- 1} [\frac{2 \sqrt{3} (x + y + 1)}{3 - 4 x y - 2 x - 2 y - 1}] = \frac{\sqrt{3} C}{2} \\ \Rightarrow \tan^{- 1} [\frac{\sqrt{3} (x + y + 1)}{2 (1 - x - y - 2 x y)}] = \frac{\sqrt{3} C}{2} \\ \Rightarrow \frac{\sqrt{3} (x + y + 1)}{2 (1 - x - y - 2 x y)} = \tan (\frac{\sqrt{3} C}{2}) = B, where B = \tan (\frac{\sqrt{3} C}{2}) \\ \Rightarrow x + y + 1 = \frac{2 B}{\sqrt{3}} (1 - x y - 2 x y) \\ \Rightarrow x + y + 1 = A (1 - x - y - 2 x y), where A = \frac{2 B}{\sqrt{3}} \end{array}$ Hence, the given result is proved.

Q8. $\begin{array}{l} Find the equation of the curve passing through the point (0, \frac{π}{4}) whose differential equation is, \\ \sin x \cos y d x + \cos x \sin y d y = 0 \end{array}$

Answer. $\begin{array}{l} The differential equation of the given curve is: \\ \sin x \cos y d x + \cos x \sin y d y = 0 \\ \Rightarrow \frac{\sin x \cos y d x + \cos x \sin y d y}{\cos x \cos y} = 0 \\ \Rightarrow \tan x d x + \tan y d y = 0 \\ \log (\sec x) + \log (\sec y) = \log C \\ \log (\sec x \cdot \sec y) = \log C \\ \Rightarrow \sec x \cdot \sec y = C . . . (i) \end{array}$ $\begin{array}{l} The curve passes through point (0, \frac{π}{4}) \\ ∴ 1 \times \sqrt{2} = C \\ \Rightarrow C = \sqrt{2} \\ On substituting C = \sqrt{2} in equation (i), we get: \\ \Rightarrow \sec x \cdot \cos y = \sqrt{2} \\ \Rightarrow \sec x \cdot \frac{1}{\sqrt{2}} \\ \Rightarrow \cos y = \frac{1}{\sqrt{2}} \end{array}$ $Hence, the required equation of the curve is \begin{array}{l} \cos y = \frac{\sec x}{\sqrt{2}} \end{array}$

Q9. $\begin{array}{l} Find the particular solution of the differential equation \\ (1 + e^{2 x}) d y + (1 + y^{2}) e^{x} d x = 0, given that y = 1 when x = 0 \end{array}$

Answer. $\begin{array}{l} (1 + e^{2 x}) d y + (1 + y^{2}) e^{x} d x = 0 \\ \Rightarrow \frac{d y}{1 + y^{2}} + \frac{e^{x} d x}{1 + e^{2 x}} = 0 \\ Integrating both sides, we get: \\ \tan^{- 1} y + \int \frac{e^{x} d x}{1 + e^{2 x}} = C . . . (i) \\ Let e^{x} = t \Rightarrow \frac{d t}{d x} \\ \Rightarrow e^{x} d x = \frac{d t}{d x} \\ \Rightarrow e^{x} d x = d t \end{array}$ $\begin{array}{l} Substituting these values in equation (i), we get: \\ \tan^{- 1} y + \int \frac{d t}{1 + t^{2}} = C \\ \Rightarrow \tan^{- 1} y + \tan^{- 1} t = C \\ \Rightarrow \tan^{- 1} y + \tan^{- 1} (e^{x}) = C . . . (i i) \end{array}$ $\begin{array}{l} Therefore, equation (i i) becomes: \\ \tan^{- 1} 1 + \tan^{- 1} 1 = C \\ \Rightarrow \frac{π}{4} + \frac{π}{4} = C \\ \Rightarrow C = \frac{π}{2} \end{array}$ $\begin{array}{l} Substituting C = \frac{π}{2} in equation (i i), we get: \\ \tan^{- 1} y + \tan^{- 1} (e^{x}) = \frac{π}{2} \end{array}$ This is the required particular solution of the given differential equation.

Q10. Solve the differential equation $y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y (y \neq 0)$

Answer. $\begin{array}{l} y e^{y} d x = (x e^{\frac{x}{y}} + y^{2}) d y \\ \Rightarrow y e^{\frac{x}{y}} \frac{d x}{d y} = x e^{\frac{x}{y}} + y^{2} \\ \Rightarrow e^{\frac{x}{y}} [y \cdot \frac{d x}{d y} - x] = y^{2} \\ \Rightarrow e^{\frac{x}{y}} \cdot \frac{[y \cdot \frac{d x}{d y} - x]}{y^{2}} = 1 . . . (i) \end{array}$ $\begin{array}{l} Let e^{\frac{x}{y}} = z \\ Differentiating it with respect to y, we get: \\ \frac{d}{d y} (e^{\frac{x}{y}}) = \frac{d z}{d y} \\ \Rightarrow e^{\frac{x}{y}} \cdot \frac{d}{d y} (\frac{x}{y}) = \frac{d z}{d y} \\ \Rightarrow e^{\frac{x}{y}} \cdot [\frac{y \cdot \frac{d x}{d y} - x}{y^{2}}] = \frac{d z}{d y} . . . (i i) \end{array}$ $\begin{array}{l} From equation (i) and equation (i i), we get: \\ \frac{d z}{d y} = 1 \\ \Rightarrow d z = d y \\ Integrating both sides, we get: \\ z = y + C \\ \Rightarrow e^{\frac{x}{y}} = y + C \end{array}$

Q11. $\begin{array}{l} Find a particular solution of the differential equation (x - y) (d x + d y) = d x - d y, given that y = - 1, \\ when x = 0 (Hint: put x - y = t) \end{array}$

Answer. $\begin{array}{l} (x - y) (d x + d y) = d x - d y \\ \Rightarrow (x - y + 1) d y = (1 - x + y) d x \\ \Rightarrow \frac{d y}{d x} = \frac{1 - x + y}{x - y + 1} \\ \Rightarrow \frac{d y}{d x} = \frac{1 - (x - y)}{1 + (x - y)} . . . (i) \end{array}$ $\begin{array}{l} Let x - y = t \\ \Rightarrow \frac{d}{d x} (x - y) = \frac{d t}{d x} \\ \Rightarrow 1 - \frac{d y}{d x} = \frac{d t}{d x} \\ \Rightarrow 1 - \frac{d t}{d x} = \frac{d y}{d x} \\ Substituting the values of x - y and \frac{d y}{d x} in equation (i), we get: \end{array}$ $\begin{array}{l} 1 - \frac{d t}{d x} = \frac{1 - t}{1 + t} \\ \Rightarrow \frac{d t}{d x} = 1 - (\frac{1 - t}{1 + t}) \\ \Rightarrow \frac{d t}{d x} = \frac{(1 + t) - (1 - t)}{1 + t} \\ \Rightarrow \frac{d t}{d x} = \frac{2 t}{1 + t} \\ \Rightarrow (\frac{1 + t}{t}) d t = 2 d x \\ \Rightarrow (1 + \frac{1}{t}) d t = 2 d x . . . (i i) \end{array}$ $\begin{array}{l} Integrating both sides, we get: \\ t + \log | t | = 2 x + C \\ \Rightarrow (x - y) + \log | x - y | = 2 x + C \\ \Rightarrow \log | x - y | = x + y + C . . . (i i i) \\ Now, y = - 1 at x = 0 \\ Therefore, equation ( iii ) becomes: \\ Tog 1 = 0 - 1 + C \\ \Rightarrow C = 1 \end{array}$ $\begin{array}{l} Substituting C = 1 in equation (i i i) we get: \\ \log | x - y | = x + y + 1 \\ This is the required particular solution of the given differential equation. \end{array}$

Q12. $\begin{array}{l} Solve the differential equation [\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}] \frac{d x}{d y} = 1 (x \neq 0) \end{array}$

Answer. $\begin{array}{l} [\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}] \frac{d x}{d y} = 1 \\ \Rightarrow \frac{d y}{d x} = \frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \\ \Rightarrow \frac{d y}{d x} + \frac{y}{\sqrt{x}} = \frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} \\ \frac{d y}{d x} + P y = Q, where P = \frac{1}{\sqrt{x}} and Q = \frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} \end{array}$ Now, I.F $= e^{\int pdx} = e^{\int \frac{1}{\sqrt{x}} d x} = e^{2 \sqrt{x}}$ $\begin{array}{l} The general solution of the given differential equation is given by, \\ y (I.F.) = \int (Q \times I.F.) d x + C \end{array}$ $\begin{array}{l} \Rightarrow y e^{2 \sqrt{x}} = \int (\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} \times e^{2 \sqrt{x}}) d x + C \\ \Rightarrow y e^{2 \sqrt{x}} = \int \frac{1}{\sqrt{x}} d x + C \\ \Rightarrow y e^{2 \sqrt{x}} = 2 \sqrt{x} + C \end{array}$

Q13. $\begin{array}{l} Find a particular solution of the differential equation \frac{d y}{d x} + y \cot x = 4 x \csc x (x \neq 0) \\ given that y = 0 when x = \frac{π}{2} \end{array}$

Answer. $\begin{array}{l} The given differential equation is: \\ \frac{d y}{d x} + y \cot x = 4 x \csc x \\ This equation is a linear differential equation of the form \\ \frac{d y}{d x} + p y = Q, where p = \cot x and Q = 4 x \csc x . \end{array}$ Now, I.F $= e^{\int p d x} = e^{\int \cot x d x} = e^{\log | \sin x |} = \sin x$ $\begin{array}{l} The general solution of the given differential equation is given by, \\ y (I.F.) = \int (Q \times I . F .) d x + C \end{array}$ $\begin{array}{l} \Rightarrow y \sin x = \int (4 x \csc x \cdot \sin x) d x + C \\ \Rightarrow y \sin x = 4 \int x d x + C \\ \Rightarrow y \sin x = 4 \cdot \frac{x^{2}}{2} + C \\ \Rightarrow y \sin x = 2 x^{2} + C . . . (i) \end{array}$ $\begin{array}{l} y = 0 at x = \frac{π}{2} \\ Therefore, equation (i) becomes: \\ 0 = 2 \times \frac{π^{2}}{4} + C \\ \Rightarrow C = - \frac{π^{2}}{2} \end{array}$ $\begin{array}{l} C = - \frac{π^{2}}{2} in equation (i), we get: \\ y \sin x = 2 x^{2} - \frac{π^{2}}{2} \\ This is the required particular solution of the given differential equation. \end{array}$

Q14. $\begin{array}{l} {Find a particular solution of the differential equation}^{(x + 1) \frac{d y}{d x} = 2 e^{- y} - 1}, given that y = 0 when \\ x = 0 \end{array}$

Answer. $\begin{array}{l} (x + 1) \frac{d y}{d x} = 2 e^{- y} - 1 \\ \Rightarrow \frac{d y}{2 e^{- y} - 1} = \frac{d x}{x + 1} \\ \Rightarrow \frac{e^{y} d y}{2 - e^{y}} = \frac{d x}{x + 1} \\ Integrating both sides, we get: \\ \int \frac{e^{y} d y}{2 - e^{y}} = \log | x + 1 | + \log C . . . (i) \end{array}$ $\begin{array}{l} Let 2 - e^{y} = t \\ ∴ \frac{d}{d y} (2 - e^{y}) = \frac{d t}{d y} \\ \Rightarrow - e^{y} = \frac{d t}{d y} \\ \Rightarrow e^{y} d t = - d t \\ Substituting this value in equation (i), we get: \\ \int \frac{- d t}{t} = \log | x + 1 | + \log C \end{array}$ $\begin{array}{l} \Rightarrow - \log | t | = \log | C (x + 1) | \\ \Rightarrow - \log | \cdot - e^{y} | = \log | C (x + 1) | \\ \Rightarrow \frac{1}{2 - e^{y}} = C (x + 1) \\ \Rightarrow 2 - e^{y} = \frac{1}{C (x + 1)} . . . (i i) \\ Now, at x = 0 and y = 0, equation (i i) becomes: \end{array}$ $\begin{array}{l} \Rightarrow 2 - 1 = \frac{1}{C} \\ \Rightarrow C = 1 \\ Substituting C = 1 in equation (i i), we get: \end{array}$ $\begin{array}{l} 2 - e^{y} = \frac{1}{x + 1} \\ \Rightarrow e^{y} = 2 - \frac{1}{x + 1} \\ \Rightarrow e^{y} = \frac{2 x + 2 - 1}{x + 1} \\ \Rightarrow e^{y} = \frac{2 x + 1}{x + 1} \\ \Rightarrow y = \log | \frac{2 x + 1}{x + 1} |, (x \neq - 1) \\ This is the required particular solution of the given differential equation. \end{array}$

Q15. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer. Let the population at any instant (t) be y. It is given that the rate of increase of population is proportional to the number of inhabitants at any instant. $\begin{array}{l} ∴ \frac{d y}{d t} \propto y \\ \Rightarrow \frac{d y}{d t} = k y (k i s c o n s t e n t) \\ \Rightarrow \frac{d y}{y} = k d t \end{array}$ Integrating both sides, we get: log y = kt + C … (i) In the year 1999, t = 0 and y = 20000. Therefore, we get: log 20000 = C … (ii) In the year 2004, t = 5 and y = 25000. Therefore, we get: $\begin{array}{l} \log 25000 = k \cdot 5 + C \\ \Rightarrow \log 25000 = 5 k + \log 20000 \\ \Rightarrow 5 k = \log (\frac{25000}{20000}) = \log (\frac{5}{4}) \\ \Rightarrow k = \frac{1}{5} \log (\frac{5}{4}) . . . (i i i) \end{array}$ $\begin{array}{l} In the year 2009, t = 10 years. \\ Now, on substituting the values of t, k, and C in equation (i), we get: \\ \log y = 10 \times \frac{1}{5} \log (\frac{5}{4}) + \log (20000) \\ \Rightarrow \log y = \log [20000 \times {(\frac{5}{4})}^{2}] \\ \Rightarrow y = 20000 \times \frac{5}{4} \times \frac{5}{4} \\ \Rightarrow y = 31250 \end{array}$ Hence, the population of the village in 2009 will be 31250.

Q16. $\begin{array}{l} The general solution of the differential equation y \frac{y d x - x d y}{y} = 0 \\ A. x y = c \\ B. x = c y^{2} \\ C. y = c x \\ D. y = c x^{2} \end{array}$

Answer. $\begin{array}{l} The given differential equation is: \\ \frac{y d x - x d y}{y} = 0 \\ \Rightarrow \frac{y d x - x d y}{x y} = 0 \\ \Rightarrow \frac{1}{x} d x - \frac{1}{y} d y = 0 \\ Integrating both sides, we get: \\ \log | x | - \log | y | = \log k \end{array}$ $\begin{array}{l} \Rightarrow \log | \frac{x}{y} | = \log k \\ \Rightarrow \frac{x}{y} = k \\ \Rightarrow y = \frac{1}{k} x \\ \Rightarrow y = C x where C = \frac{1}{k} \\ Hence, the correct answer is C . \end{array}$

Q17. The general solution of a differential equation of the type $\frac{d x}{d y} + P_{1} x = Q_{1}$ is $A . y e^{\int p_{1} d y} = \int (Q_{1} e^{\int p_{1} d y}) d y + C$ $B \cdot e^{\int P_{1} d x}$ $= \int (Q_{l} e^{\int p_{1} d x}) d x + C$ $C . x e^{\int p_{1} d y} = \int (Q_{1} e^{\int p_{1} d y}) d y + C$ $D . x e^{\int p_{d x}} = \int (Q_{1} e^{\int p_{1} d x}) d x + C$

Answer. The integrating factor of the given differential equation $\frac{d x}{d y} + P_{1} x = Q_{1} is e^{\int P_{1 d y}} .$ The general solution of the differential equation is given by, $x (I.F.) = \int (Q \times I . F .) d y + C$ Solutions Class 12 Maths Chapter-9 (Determinants)Miscellaneous Exercise Hence, the correct answer is C.

Q18. $\begin{array}{l} The general solution of the differential equation e^{x} d y + (y e^{x} + 2 x) d x = 0 is \\ A. x e^{y} + x^{2} = c \\ B. x e^{y} + y^{2} = c \\ C. y e^{x} + x^{2} = c \\ D. y e^{y} + x^{2} = c \end{array}$

Answer. $\begin{array}{l} The given differential equation is: \\ e^{x} d y + (y e^{x} + 2 x) d x = 0 \\ \Rightarrow e^{x} \frac{d y}{d x} + y e^{x} + 2 x = 0 \\ \Rightarrow \frac{d y}{d x} + y = - 2 x e^{- x} \\ This is a linear differential equation of the form \\ \frac{d y}{d x} + P y = Q, where P = 1 and Q = - 2 x e^{- x} . \end{array}$ Now, I.F $= e^{\int p d x} = e^{\int d x} = e^{x}$ The general solution of the given differential equation is given by, $y (I.F.) = \int (Q \times I . F .) d x + C$ $\begin{array}{l} \Rightarrow y e^{x} = \int (- 2 x e^{- x} \cdot e^{x}) d x + C \\ \Rightarrow y e^{x} = - \int 2 x d x + C \\ \Rightarrow y e^{x} = - x^{2} + C \\ \Rightarrow y e^{x} + x^{2} = C \\ Hence, the correct answer is C . \end{array}$