Question 18:

The tangent at any point (x, y) of a curve makes an angle tan⁻¹(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).

Answer:

The slope of the curve is given as $\frac{d y}{d x} = \tan θ$ .
Here,

$θ = \tan^{- 1} (2 x + 3 y) ∴ \frac{d y}{d x} = \tan (\tan^{- 1} 2 x + 3 y) \Rightarrow \frac{d y}{d x} = 2 x + 3 y$

$\Rightarrow \frac{d y}{d x} - 3 y = 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 3 and Q = 2 x ∴ I . F . = e^{\int P d x} = e^{\int - 3 d x} = e^{- 3 x} Multiplying both sides of (1), by I . F . = e^{- 3 x}, we get e^{- 3 x} (\frac{d y}{d x} - 3 y) = e^{- 3 x} . 2 x \Rightarrow e^{- 3 x} (\frac{d y}{d x} - 3 y) = 2 x e^{- 3 x} Integrating both sides with respect to x, we get y e^{- 3 x} = 2 \int x e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = 2 \int \underset{I}{x} \underset{II}{e^{- 3 x}} d x + C \Rightarrow y e^{- 3 x} = 2 x \int e^{- 3 x} d x - 2 \int [\frac{d}{d x} (x) \int e^{- 3 x} d x] d x + C \Rightarrow y e^{- 3 x} = - 2 x \frac{e^{- 3 x}}{3} + 2 \times \frac{1}{3} \int e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = - \frac{2}{3} x e^{- 3 x} - 2 \times \frac{1}{9} e^{- 3 x} + C \Rightarrow y e^{- 3 x} = - \frac{2}{3} x e^{- 3 x} - \frac{2}{9} e^{- 3 x} + C Since the curve passes through (1, 2), it satisfies the above equation . ∴ 2 e^{- 3} = - \frac{2}{3} e^{- 3} - \frac{2}{9} e^{- 3} + C \Rightarrow C = 2 e^{- 3} + \frac{2}{3} e^{- 3} + \frac{2}{9} e^{- 3} \Rightarrow C = \frac{26}{9} e^{- 3} Putting the value of C, we get y e^{- 3 x} = (- \frac{2}{3} x - \frac{2}{9}) e^{- 3 x} + \frac{26}{9} e^{- 3}$

Page No 21.135:

Question 19:

Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).

Answer:

Portion of the x-axis cut off between the origin and tangent at a point $= x - y \frac{d x}{d y} = OT$

It is given, OT = 2x

$\begin{array}{l} ∴ x - y \frac{d x}{d y} = 2 x \\ - x = y \frac{d x}{d y} \\ - \int \frac{d x}{x} = \int \frac{d y}{y} \\ ∴ x y = k \end{array}$

Since the curve passes through the point (1, 2)
⇒ at x = 1 ⇒ y = 2
∴ k = 2
∴ xy = 2

$(i) {(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0$
The highest order derivative in the given equation is $\frac{d^{2} s}{d t^{2}}$ and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(ii) y"' + 2y" + y' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1

(iii) (y"')² + (y")³ + (y')⁴ + y⁵ = 0
The highest order derivative in the given equation is y''' and its power is 2.
Therefore, the given differential equation is of third order and second degree.
i.e., Order = 3 and degree = 2

(iv) y"' + 2y" + y' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1

(v) y" + (y')² + 2y = 0
The highest order derivative in the given equation is y'' and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(vi) y" + 2y' + sin y = 0
The highest order derivative in the given equation is y'' and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(vii) y"' + y² + e^y^' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order. This equation cannot be expressed as a polynomial of derivative.
Thus, the degree is not defined.
i.e., Order = 3 and degree is not defined.

Page No 21.138:

Question 2:

Verify that the function y = e^−3x is a solution of the differential equation $\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - 6 y = 0$ .

Answer:

^{$We have, \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - 6 y = 0 . . . . . (1) Now, y = e^{- 3 x} \Rightarrow \frac{d y}{d x} = - 3 e^{- 3 x} \Rightarrow \frac{d^{2} y}{d x^{2}} = 9 e^{- 3 x}$}
$Putting the values of \frac{d^{2} y}{d x^{2}}, \frac{d y}{d x} and y in (1), we get$

$LHS = 9 e^{- 3 x} - 3 e^{- 3 x} - 6 e^{- 3 x} = 0 = RHS$

Thus, y = e^−3x is the solution of the given differential equation.

Page No 21.138:

Question 3:

In each of the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

(i) y = e^x + 1	y'' − y' = 0
(ii) y = x² + 2x + C	y' − 2x − 2 = 0
(iii) y = cos x + C	y' + sin x = 0
(iv) y = $\sqrt{1 + x^{2}}$	y' = $\frac{x y}{1 + x^{2}}$
(v) y = x sin x	xy' = y + x $\sqrt{x^{2} - y^{2}}$
(vi) $y = \sqrt{a^{2} - x^{2}}$	$x + y \frac{d y}{d x} = 0$

Answer:

$(i) We have, x (x^{2} - 1) \frac{d y}{d x} = 1 \Rightarrow \frac{d y}{d x} = \frac{1}{x (x^{2} - 1)} \Rightarrow d y = \{\frac{1}{x (x^{2} - 1)}\} d x Integrating both sides, we get \int d y = \int \{\frac{1}{x (x^{2} - 1)}\} d x \Rightarrow y = \int \{\frac{1}{x (x^{2} - 1)}\} d x + C \Rightarrow y = \int \{\frac{1}{x (x + 1) (x - 1)}\} d x + C . . . . . (1) Let \frac{1}{x (x + 1) (x - 1)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1} \Rightarrow 1 = A (x + 1) (x - 1) + B x (x - 1) + C x (x + 1) \Rightarrow 1 = A (x^{2} - 1) + B (x^{2} - x) + C (x^{2} + x) \Rightarrow 1 = x^{2} (A + B + C) + x (- B + C) - A Comparing both sides, we get - A = 1 . . . . . (2) - B + C = 0 . . . . . (3) A + B + C = 0 . . . . . (4) Solving (2), (3) and (4), we get A = - 1 B = \frac{1}{2} C = \frac{1}{2} ∴ \frac{1}{x (x + 1) (x - 1)} = \frac{- 1}{x} + \frac{1}{2 (x + 1)} + \frac{1}{2 (x - 1)} Now, (1) becomes y = \int \{\frac{- 1}{x} + \frac{1}{2 (x + 1)} + \frac{1}{2 (x - 1)}\} d x + C \Rightarrow y = - \int \frac{1}{x} d x + \frac{1}{2} \int \frac{1}{x - 1} d x + \frac{1}{2} \int \frac{1}{x - 1} d x \Rightarrow y = - \log |x| + \frac{1}{2} \log |x - 1| + \frac{1}{2} \log |x + 1| + C \Rightarrow y = \frac{1}{2} \log |x - 1| + \frac{1}{2} \log |x + 1| - \log |x| + C Given : y (2) = 0 ∴ 0 = \frac{1}{2} \log |2 - 1| + \frac{1}{2} \log |2 + 1| - \log |2| + C \Rightarrow C = \log |2| - \frac{1}{2} \log |3| Substituting the value of C, we get y = \frac{1}{2} \log |x - 1| + \frac{1}{2} \log |x + 1| - \log |x| + \log |2| - \frac{1}{2} \log |3| \Rightarrow 2 y = \log |x - 1| + \log |x + 1| - 2 \log |x| + 2 \log |2| - \log |3| \Rightarrow 2 y = \log |x - 1| + \log |x + 1| - \log |x^{2}| + \log |4| - \log |3| \Rightarrow 2 y = \log \frac{(x - 1) (x + 1)}{x^{2}} - (\log |3| - \log |4|) \Rightarrow y = \frac{1}{2} \log \frac{(x^{2} - 1)}{x^{2}} - \frac{1}{2} \log (\frac{3}{4})$

$(ii) We have, \cos (\frac{d y}{d x}) = a \Rightarrow \frac{d y}{d x} = \cos^{- 1} a \Rightarrow d y = \cos^{- 1} a d x Integrating both sides, we get \int d y = \int \cos^{- 1} a d x \Rightarrow y = x \cos^{- 1} a + C Now, When x = 0, y = 1 ∴ 1 = 0 + C \Rightarrow C = 1 Putting the value of C in (1), we get y = x \cos^{- 1} a + 1 \Rightarrow \cos (\frac{y - 1}{x}) = a$

$(iii) We have, \frac{d y}{d x} = y \tan x \Rightarrow \frac{1}{y} d y = \tan x d x Integrating both sides, we get \int \frac{1}{y} d y = \int \tan x d x \Rightarrow \log y = \log |\sec x| + C . . . . (1) Now, When x = 0, y = 1 ∴ \log 1 = \log 1 + C \Rightarrow C = 0 Putting the value of C in (1), we get \log y = \log |\sec x| \Rightarrow y = \sec x$

Page No 21.140:

Question 66:

Solve the each of the following differential equations:
(i) $(x - y) \frac{d y}{d x} = x + 2 y$

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

(iii) y dx + x log $(\frac{y}{x})$ dy − 2x dy = 0

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

(v) $x \frac{d y}{d x} + 2 y = x^{2}, x \neq 0$

(vi) $\frac{d y}{d x} + 2 y = \sin x$

(vii) $\frac{d y}{d x} + 3 y = e^{- 2 x}$

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

(ix) $\frac{d y}{d x} + (\sec x) y = \tan x$

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

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

(xii) (1 + x²) dy + 2xy dx = cot x dx

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

(xiv) y dx + (x − y²) dy = 0

(xv) $(x + 3 y^{2}) \frac{d y}{d x} = y$

Answer:

$(i) We have, (x - y) \frac{d y}{d x} = x + 2 y \Rightarrow \frac{d y}{d x} = \frac{x + 2 y}{x - y} . . . . . (1) Clearly this is a homogeneous equation, Putting y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} Substituting y = v x and \frac{d y}{d x} = v + x \frac{d v}{d x} (1) becomes, v + x \frac{d v}{d x} = \frac{x + 2 v x}{x - v x} \Rightarrow v + x \frac{d v}{d x} = \frac{1 + 2 v}{1 - v} \Rightarrow x \frac{d v}{d x} = \frac{1 + 2 v}{1 - v} - v \Rightarrow x \frac{d v}{d x} = \frac{1 + 2 v - v + v^{2}}{1 - v} \Rightarrow x \frac{d v}{d x} = \frac{v^{2} + v + 1}{1 - v} \Rightarrow \frac{1 - v}{v^{2} + v + 1} d v = \frac{1}{x} d x \Rightarrow [\frac{- v}{v^{2} + v + 1} + \frac{1}{v^{2} + v + 1}] d v = \frac{1}{x} d x \Rightarrow [- \frac{1}{2} \times \frac{2 v + 1 - 1}{v^{2} + v + 1} + \frac{1}{v^{2} + v + 1}] d v = \frac{1}{x} d x \Rightarrow [- \frac{1}{2} \times \frac{2 v + 1}{v^{2} + v + 1} + \frac{1}{2} \times \frac{1}{v^{2} + v + 1} + \frac{1}{v^{2} + v + 1}] d v = \frac{1}{x} d x \Rightarrow [- \frac{1}{2} \times \frac{2 v + 1}{v^{2} + v + 1} + \frac{3}{2} \times \frac{1}{v^{2} + v + 1}] d v = \frac{1}{x} d x \Rightarrow [- \frac{1}{2} \times \frac{2 v + 1}{v^{2} + v + 1} + \frac{3}{2} \times \frac{1}{v^{2} + v + \frac{1}{4} + \frac{3}{4}}] d v = \frac{1}{x} d x \Rightarrow [- \frac{1}{2} \times \frac{2 v + 1}{v^{2} + v + 1} + \frac{3}{2} \times \frac{1}{{(v + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}] d v = \frac{1}{x} d x Integrating both sides, we get \Rightarrow \int [- \frac{1}{2} \times \frac{2 v + 1}{v^{2} + v + 1} + \frac{3}{2} \times \frac{1}{{(v + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}] d v = \int \frac{1}{x} d x \Rightarrow - \frac{1}{2} \int \frac{2 v + 1}{v^{2} + v + 1} d v + \frac{3}{2} \int \frac{1}{{(v + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d v = \int \frac{1}{x} d x \Rightarrow - \frac{1}{2} \log |v^{2} + v + 1| + \frac{3}{2} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \log |x| + C \Rightarrow - \frac{1}{2} \log |{(\frac{y}{x})}^{2} + \frac{y}{x} + 1| + \frac{3}{2} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} \frac{\frac{y}{x} + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \log |x| + C \Rightarrow - \frac{1}{2} \log |\frac{y^{2} + x y + x^{2}}{x^{2}}| + \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) = \log |x| + C \Rightarrow - \frac{1}{2} \log |y^{2} + x y + x^{2}| + \frac{1}{2} \log |x^{2}| + \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) = \log |x| + C \Rightarrow - \frac{1}{2} \log |y^{2} + x y + x^{2}| + \log |x| + \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) = \log |x| + C \Rightarrow - \frac{1}{2} \log |y^{2} + x y + x^{2}| + \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) = C \Rightarrow \log |y^{2} + x y + x^{2}| - 2 \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) = - 2 C \Rightarrow \log |y^{2} + x y + x^{2}| = 2 \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) - 2 C \Rightarrow \log |y^{2} + x y + x^{2}| = 2 \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3} x}) + k Where, k = - 2 C$

$(ii) We have, x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x \Rightarrow \frac{d y}{d x} = \frac{y \cos (\frac{y}{x}) + x}{x \cos (\frac{y}{x})} . . . . . (1) Clearly this is a homogeneous equation, Putting y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} Substituting y = v x and \frac{d y}{d x} = v + x \frac{d v}{d x} in (1) we get \frac{d y}{d x} = \frac{y \cos (\frac{y}{x}) + x}{x \cos (\frac{y}{x})} \Rightarrow v + x \frac{d v}{d x} = \frac{v x \cos (v) + x}{x \cos (v)} \Rightarrow v + x \frac{d v}{d x} = \frac{v \cos (v) + 1}{\cos (v)} \Rightarrow x \frac{d v}{d x} = \frac{v \cos (v) + 1}{\cos (v)} - v \Rightarrow x \frac{d v}{d x} = \frac{v \cos (v) + 1 - v \cos (v)}{\cos (v)} \Rightarrow x \frac{d v}{d x} = \frac{1}{\cos (v)} \Rightarrow \cos (v) d v = \frac{1}{x} d x Integrating both sides, we get \int \cos (v) d v = \int \frac{1}{x} d x \Rightarrow \sin (v) = \log |x| + \log |C| \Rightarrow \sin (\frac{y}{x}) = \log |C x|$

$(iii) We have, y d x + x \log (\frac{y}{x}) d y - 2 x d y = 0 \Rightarrow x \log (\frac{y}{x}) d y - 2 x d y = - y d x \Rightarrow [\log (\frac{y}{x}) - 2] x d y = - y d x \Rightarrow \frac{d y}{d x} = \frac{- y}{[\log (\frac{y}{x}) - 2] x} \Rightarrow \frac{d y}{d x} = \frac{\frac{y}{x}}{2 - \log (\frac{y}{x})} . . . . . (1) Clearly this is a homogenous equation, Putting y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} Substituting y = v x and \frac{d y}{d x} = v + x \frac{d v}{d x} in (1) we get v + x \frac{d v}{d x} = \frac{v}{2 - \log (v)} \Rightarrow x \frac{d v}{d x} = \frac{v}{2 - \log (v)} - v \Rightarrow x \frac{d v}{d x} = \frac{v - 2 v + v \log (v)}{2 - \log (v)} \Rightarrow x \frac{d v}{d x} = \frac{- v + v \log (v)}{2 - \log (v)} \Rightarrow \frac{2 - \log (v)}{- v + v \log (v)} d v = \frac{1}{x} d x \Rightarrow \frac{\log (v) - 2}{v \log (v) - v} d v = - \frac{1}{x} d x \Rightarrow \frac{\log (v) - 1 - 1}{v [\log (v) - 1]} d v = - \frac{1}{x} d x \Rightarrow \frac{\log (v) - 1}{v [\log (v) - 1]} d v - \frac{1}{v [\log (v) - 1]} d v = - \frac{1}{x} d x \Rightarrow \frac{1}{v} d v - \frac{1}{v [\log (v) - 1]} d v = - \frac{1}{x} d x Integrating both sides we get \int \frac{1}{v} d v - \int \frac{1}{v [\log (v) - 1]} d v = - \int \frac{1}{x} d x \Rightarrow \log |v| - I = - \log |x| - \log C . . . . . (2) Where, I = \int \frac{1}{v [\log |(v)| - 1]} d v Puting \log v = t \frac{1}{v} d v = d t ∴ I = \int \frac{1}{t - 1} d t \Rightarrow I = \log |t - 1| \Rightarrow I = \log |\log |v| - 1| . . . . . (3) From (2) and (3) we get \log |v| - \log |\log |v| - 1| = - \log |x| - \log C \Rightarrow \log |\frac{v}{\log |v| - 1}| = - \log |C x| \Rightarrow \frac{v}{\log |v| - 1} = \frac{1}{C x} \Rightarrow \log |v| - 1 = v C x \Rightarrow \log |\frac{y}{x}| - 1 = C y$

$(iv) We have, \frac{d y}{d x} - y = \cos x Comparing with \frac{d y}{d x} + P y = Q, we get P = - 1 Q = \cos x Now, I . F . = e^{- 1 \int d x} = e^{- x} Solution is given by, y \times I . F . = \int \cos x \times I . F . d x + C \Rightarrow y e^{- x} = \int e^{- x} \cos x d x + C \Rightarrow y e^{- x} = I + C . . . . . (1) Where, I = \int \underset{II}{e^{- x}} \underset{I}{\cos x} d x . . . . . (2) \Rightarrow I = \cos x \int e^{- x} d x - \int [\frac{d}{d x} (\cos x) \int e^{- x} d x] d x \Rightarrow I = - \cos x e^{- x} - \int \sin x e^{- x} d x \Rightarrow I = - \cos x e^{- x} - \int \underset{I}{\sin x} \underset{II}{e^{- x}} d x \Rightarrow I = - \cos x e^{- x} - \sin x \int e^{- x} d x + \int [\frac{d}{d x} (\sin x) \int e^{- x} d x] d x \Rightarrow I = - \cos x e^{- x} +$