Find one-parameter families of solution curves of the following differential equations-or Solve the following differential equationsi dydx -3 y - emx - m is a given real numberii d y d x-y-cos 2 xiii x d y d x-y-x-1 e-xiv x d y d x-y-x 4v x log x d y d x-y-log xvi d y d x-2 x y 1-x 2-x 2-2vii d y d x-y cos x-esin x cos xviii x-y d y d x-1ix d y d x cos 2 x-tan x-yx e-y sec2 y d y-d x-x d yxi x log x d y d x-y-2 log xxii x d y d x-2 y-x 2 log x

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Byju's Answer

Standard XII

Mathematics

Eccentricity of Hyperbola

Find one-para...

Question

Find one-parameter families of solution curves of the following differential equations:
(or Solve the following differential equations)
(i) $\frac{d y}{d x} + 3 y = e^{m x}$ , m is a given real number

(ii) $\frac{d y}{d x} - y = \cos 2 x$

(iii) $x \frac{d y}{d x} - y = (x + 1) e^{- x}$

(iv) $x \frac{d y}{d x} + y = x^{4}$

(v) $(x \log x) \frac{d y}{d x} + y = \log x$

(vi) $\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = x^{2} + 2$

(vii) $\frac{d y}{d x} + y \cos x = e^{\sin x} \cos x$

(viii) $(x + y) \frac{d y}{d x} = 1$

(ix) $\frac{d y}{d x} \cos^{2} x = \tan x - y$

(x) $e^{- y} \sec^{2} y d y = d x + x d y$

(xi) $x \log x \frac{d y}{d x} + y = 2 \log x$

(xii) $x \frac{d y}{d x} + 2 y = x^{2} \log x$

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Solution

$(i) We have, \frac{d y}{d x} + 3 y = e^{m x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = 3 Q = e^{m x} ∴ I . F . = e^{\int P d x} = e^{\int 3 d x} = e^{3 x} Multiplying both sides of (1) by e^{3 x}, we get e^{3 x} (\frac{d y}{d x} + 3 y) = e^{3 x} e^{m x} \Rightarrow e^{3 x} \frac{d y}{d x} + 3 e^{3 x} y = e^{(m + 3) x} Integrating both sides with respect to x, we get y e^{3 x} = \int e^{(m + 3) x} d x + C (when m + 3 \neq 0) \Rightarrow y e^{3 x} = \frac{e^{(m + 3) x}}{m + 3} + C \Rightarrow y = \frac{e^{m x}}{m + 3} + C e^{- 3 x} y e^{3 x} = \int e^{0 \times x} d x + C (when m + 3 = 0) \Rightarrow y e^{3 x} = \int d x + C \Rightarrow y e^{3 x} = x + C \Rightarrow y = (x + C) e^{- 3 x} Hence, y = \frac{e^{m x}}{m + 3} + C e^{- 3 x}, where m + 3 \neq 0 and y = (x + C) e^{- 3 x}, where m + 3 = 0 are required solutions .$

$(ii) We have, \frac{d y}{d x} - y = \cos 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 1 Q = \cos 2 x ∴ I . F . = e^{\int P d x} = e^{- \int d x} = e^{- x} Multiplying both sides of (1) by e^{- x}, we get e^{- x} (\frac{d y}{d x} - y) = e^{- x} \cos 2 x \Rightarrow e^{- x} \frac{d y}{d x} - e^{- x} y = e^{- x} \cos 2 x Integrating both sides with respect to x, we get y e^{- x} = \int e^{- x} \cos 2 x d x + C \Rightarrow y e^{- x} = I + C . . . . . (2) Where, I = \int e^{- x} \cos 2 x d x . . . . . (3) \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{2} \int (- e^{- x} \sin 2 x) d x \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x + \frac{1}{2} \int e^{- x} \sin 2 x d x \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{4} e^{- x} \cos 2 x - \frac{1}{2} \times \frac{1}{2} \int [(- e^{- x}) \times (- \cos 2 x)] d x \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{4} e^{- x} \cos 2 x - \frac{1}{4} \int e^{- x} \cos 2 x d x \Rightarrow I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{4} e^{- x} \cos 2 x - \frac{1}{4} I [From (3)] \Rightarrow \frac{5}{4} I = \frac{1}{2} e^{- x} \sin 2 x - \frac{1}{4} e^{- x} \cos 2 x \Rightarrow 5 I = 2 e^{- x} \sin 2 x - e^{- x} \cos 2 x \Rightarrow I = \frac{e^{- x}}{5} (2 \sin 2 x - \cos 2 x) . . . . . (4) From (2) and (4) we get \Rightarrow y e^{- x} = \frac{e^{- x}}{5} (2 \sin 2 x - \cos 2 x) + C \Rightarrow y = \frac{1}{5} (2 \sin 2 x - \cos 2 x) + C e^{x} Hence, y = \frac{1}{5} (2 \sin 2 x - \cos 2 x) + C e^{x} is the required solution .$

$(iii) We have, x \frac{d y}{d x} - y = (x + 1) e^{- x} \Rightarrow \frac{d y}{d x} - \frac{1}{x} y = (\frac{x + 1}{x}) e^{- x} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{1}{x} Q = (\frac{x + 1}{x}) e^{- x} ∴ I . F . = e^{\int P d x} = e^{- \int \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x} Multiplying both sides of (1) by \frac{1}{x}, we get \frac{1}{x} (\frac{d y}{d x} - \frac{1}{x} y) = \frac{1}{x} (\frac{x + 1}{x}) e^{- x} \Rightarrow \frac{1}{x} \frac{d y}{d x} - \frac{1}{x^{2}} y = (\frac{x + 1}{x^{2}}) e^{- x} Integrating both sides with respect to x, we get \frac{1}{x} y = \int (\frac{1}{x} + \frac{1}{x^{2}}) e^{- x} d x + C . . . . . (2) Putting \frac{1}{x} e^{- x} = t \Rightarrow (- \frac{1}{x} e^{- x} - \frac{1}{x^{2}} e^{- x}) d x = d t \Rightarrow (\frac{1}{x} + \frac{1}{x^{2}}) e^{- x} d x = - d t Therefore (2) becomes \frac{1}{x} y = - \int d t + C \Rightarrow \frac{1}{x} y = - t + C \Rightarrow \frac{1}{x} y = - \frac{1}{x} e^{- x} + C \Rightarrow y = - e^{- x} + C x Hence, y = - e^{- x} + C x is the required solution .$

$(iv) We have, x \frac{d y}{d x} + y = x^{4} \Rightarrow \frac{d y}{d x} + \frac{1}{x} y = x^{3} . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \frac{1}{x} Q = x^{3} ∴ I . F . = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{\log x} = x Multiplying both sides of (1) by x, we get x (\frac{d y}{d x} + \frac{1}{x} y) = x . x^{3} \Rightarrow x \frac{d y}{d x} + y = x^{4} Integrating both sides with respect to x, we get x y = \int x^{4} d x + C \Rightarrow x y = \frac{x^{5}}{5} + C \Rightarrow y = \frac{x^{4}}{5} + \frac{C}{x} Hence, y = \frac{x^{4}}{5} + \frac{C}{x} is the required solution .$

$(v) We have, (x \log x) \frac{d y}{d x} + y = \log x Dividing both sides by x \log x, we get \frac{d y}{d x} + \frac{y}{x \log x} = \frac{\log x}{x \log x} \Rightarrow \frac{d y}{d x} + \frac{y}{x \log x} = \frac{1}{x} \Rightarrow \frac{d y}{d x} + (\frac{1}{x \log x}) y = \frac{1}{x} Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{1}{x \log x} Q = \frac{1}{x} Now, I . F . = e^{\int P d x} = e^{\int \frac{1}{x \log x} d x} = e^{\log (\log x)} = \log x So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow y \log x = \int \frac{1}{x} \times \log x d x + C \Rightarrow y \log x = \frac{{(\log x)}^{2}}{2} + C \Rightarrow y = \frac{1}{2} \log x + \frac{C}{\log x}$

$(vi) We have, \frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = x^{2} + 2 . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{2 x}{1 + x^{2}} Q = x^{2} + 2 ∴ I . F . = e^{\int P d x} = e^{- \int \frac{2 x}{1 + x^{2}} d x} = e^{- \log |1 + x^{2}|} = \frac{1}{1 + x^{2}} Multiplying both sides of (1) by \frac{1}{1 + x^{2}}, we get \frac{1}{1 + x^{2}} (\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}}) = \frac{1}{1 + x^{2}} (x^{2} + 2) \Rightarrow \frac{1}{1 + x^{2}} \frac{d y}{d x} - \frac{2 x y}{{(1 + x^{2})}^{2}} = \frac{x^{2} + 2}{x^{2} + 1} Integrating both sides with respect to x, we get \frac{1}{1 + x^{2}} y = \int \frac{x^{2} + 2}{x^{2} + 1} d x + C \Rightarrow \frac{1}{1 + x^{2}} y = \int \frac{x^{2} + 1 + 1}{x^{2} + 1} d x + C \Rightarrow \frac{1}{1 + x^{2}} y = \int d x + \int \frac{1}{x^{2} + 1} d x + C \Rightarrow \frac{1}{1 + x^{2}} y = x + \tan^{- 1} x + C \Rightarrow y = (1 + x^{2}) (x + \tan^{- 1} x + C) Hence, y = (1 + x^{2}) (x + \tan^{- 1} x + C) is the required solution .$

$(vii) We have, \frac{d y}{d x} + y \cos x = e^{\sin x} \cos x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \cos x Q = e^{\sin x} \cos x ∴ I . F . = e^{\int P d x} = e^{\int \cos x d x} = e^{\sin x} Multiplying both sides of (1) by e^{\sin x}, we get e^{\sin x} (\frac{d y}{d x} + y \cos x) = e^{\sin x} \times e^{\sin x} \cos x \Rightarrow e^{\sin x} \frac{d y}{d x} + y e^{\sin x} \cos x = e^{2 \sin x} \cos x Integrating both sides with respect to x, we get e^{\sin x} y = \int e^{2 \sin x} \cos x d x + C \Rightarrow e^{\sin x} y = I + C . . . . . (2) Where, I = \int e^{2 \sin x} \cos x d x Putting t = \sin x, we get d t = \cos x d x ∴ I = \int e^{2 t} d t = \frac{e^{2 t}}{2} = \frac{e^{2 \sin x}}{2} Putting the value of I in (2), we get e^{\sin x} y = \frac{e^{2 \sin x}}{2} + C \Rightarrow y = \frac{e^{\sin x}}{2} + C e^{- \sin x} Hence, y = \frac{e^{\sin x}}{2} + C e^{- \sin x} is the required solution .$

$(viii) We have, (x + y) \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{x + y} \Rightarrow \frac{d x}{d y} = x + y \Rightarrow \frac{d x}{d y} - x = y . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = - 1 Q = y ∴ I . F . = e^{\int P d y} = e^{\int - 1 d y} = e^{- y} Multiplying both sides of (1) by e^{- y}, we get e^{- y} (\frac{d x}{d y} - x) = e^{- y} y \Rightarrow e^{- y} \frac{d x}{d y} - e^{- y} x = e^{- y} y Integrating both sides with respect to y, we get e^{- y} x = \int \underset{I}{y} \underset{II}{e^{- y}} d y + C \Rightarrow e^{- y} x = y \int e^{- y} d y - \int [\frac{d}{d y} (y) \int e^{- y} d y] d y + C \Rightarrow e^{- y} x = - y e^{- y} - e^{- y} + C \Rightarrow e^{- y} x + y e^{- y} + e^{- y} = C \Rightarrow (x + y + 1) e^{- y} = C \Rightarrow (x + y + 1) = C e^{y} Hence, (x + y + 1) = C e^{y} is the required solution .$

$(ix) We have, \frac{d y}{d x} \cos^{2} x = \tan x - y \Rightarrow \frac{d y}{d x} + \frac{1}{\cos^{2} x} y = \tan x \sec^{2} x \Rightarrow \frac{d y}{d x} + y \sec^{2} x = \tan x \sec^{2} x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = \sec^{2} x Q = \tan x \sec^{2} x ∴ I . F . = e^{\int P d x} = e^{\int \sec^{2} x d x} = e^{\tan x} Multiplying both sides of (1) by e^{\tan x}, we get e^{\tan x} (\frac{d y}{d x} + y \sec^{2} x) = e^{\tan x} \tan x \sec^{2} x \Rightarrow e^{\tan x} \frac{d y}{d x} + y e^{\tan x} \sec^{2} x = e^{\tan x} \tan x \sec^{2} x Integrating both sides with respect to x, we get e^{\tan x} y = \int e^{\tan x} \tan x \sec^{2} x d x + C \Rightarrow e^{\tan x} y = I + C . . . . . (2) Where, I = \int e^{\tan x} \tan x \sec^{2} x d x Putting t = \tan x, we get d t = \sec^{2} x d x ∴ I = \int \underset{I}{t} \underset{II}{e^{t}} d t = t \int e^{t} d t - \int [\frac{d}{d t} (t) \int e^{t} d t] d t = t e^{t} - e^{t} = (t - 1) e^{t} = (\tan x - 1) e^{\tan x} Putting the value of I in (2), we get e^{\tan x} y = (\tan x - 1) e^{\tan x} + C \Rightarrow y = (\tan x - 1) + C e^{- \tan x} Hence, y = (\tan x - 1) + C e^{- \tan x} is the required solution .$

$(x) We have, e^{- y} \sec^{2} y d y = d x + x d y \Rightarrow d x = e^{- y} \sec^{2} y d y - x d y \Rightarrow \frac{d x}{d y} = e^{- y} \sec^{2} y - x \Rightarrow \frac{d x}{d y} + x = e^{- y} \sec^{2} y . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d x}{d y} + P x = Q where P = 1 Q = e^{- y} \sec^{2} y ∴ I . F . = e^{\int P d y} = e^{\int d y} = e^{y} Multiplying both sides of (1) by e^{y}, we get e^{y} (\frac{d x}{d y} + x) = e^{y} e^{- y} \sec^{2} y \Rightarrow e^{y} \frac{d x}{d y} + e^{y} x = \sec^{2} y Integrating both sides with respect to y, we get e^{y} x = \int \sec^{2} y d y + C \Rightarrow e^{y} x = \tan y + C \Rightarrow x = (\tan y + C) e^{- y} Hence, x = (\tan y + C) e^{- y} is the required solution .$

$(xi) We have, x \log x \frac{d y}{d x} + y = 2 \log x Dividing both sides by x \log x, we get \frac{d y}{d x} + \frac{y}{x \log x} = 2 \frac{\log x}{x \log x} \Rightarrow \frac{d y}{d x} + \frac{y}{x \log x} = \frac{2}{x} \Rightarrow \frac{d y}{d x} + (\frac{1}{x \log x}) y = \frac{2}{x} Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{1}{x \log x} Q = \frac{2}{x} Now, I . F . = e^{\int P d x} = e^{\int \frac{1}{x \log x} d x} = e^{\log |\log x|} = \log x So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow y \log x = 2 \int \frac{1}{x} \times \log x d x + C Putting \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ y \log x = 2 \int t d t + C \Rightarrow y \log x = \frac{2 t^{2}}{2} + C \Rightarrow y \log x = t^{2} + C \Rightarrow y \log x = {(\log x)}^{2} + C (∵ \log x = t) \Rightarrow y = \log x + \frac{C}{\log x}$

$(xii) We have, x \frac{d y}{d x} + 2 y = x^{2} \log x Dividing both sides by x, we get \frac{d y}{d x} + \frac{2 y}{x} = x \log x Comparing with \frac{d y}{d x} + P y = Q, we get P = \frac{2}{x} Q = x \log x Now, I . F . = e^{\int P d x} = e^{\int \frac{2}{x} d x} = e^{2 \log |x|} = x^{2} So, the solution is given by y \times I . F . = \int Q \times I . F . d x + C \Rightarrow x^{2} y = \int \underset{II}{x^{3}} \underset{I}{\log x} d x + C \Rightarrow x^{2} y = \log x \int x^{3} d x - \int [\frac{d}{d x} (\log x) \int x^{3} d x] d x + C \Rightarrow x^{2} y = \frac{x^{4} \log x}{4} - \int \frac{x^{3}}{4} d x + C \Rightarrow x^{2} y = \frac{x^{4} \log x}{4} - \frac{x^{4}}{16} + C \Rightarrow y = \frac{x^{2} \log x}{4} - \frac{x^{2}}{16} + \frac{C}{x^{2}} \Rightarrow y = \frac{x^{2}}{16} (4 \log x - 1) + \frac{C}{x^{2}}$

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