Differential Equations JEE Mathematics Worksheets Set 02

Subjective Questions

Question. Show that the curve such that the distance between the origin and the tangent at an arbitrary point is equal to the distance between the origin and the normal at the same point. \( \sqrt{x^2 + y^2} = ce^{\pm \tan^{-1} \frac{y}{x}} \)
Answer:
tangent \( Y - y = m (X - x) \)
Normal \( Y - y = \frac{-1}{m} (X - x) \)
\( \perp^r \) from origin to tangent = \( \perp^r \) to Normal
\( \left| \frac{-y + mx}{\sqrt{1 + m^2}} \right| = \left| \frac{-y - x/m}{\sqrt{1 + 1/m^2}} \right| \)
\( (y - mx)^2 = (my + x)^2 \)
\( \implies \) \( m^2 (x^2 - y^2) - 4mxy + y^2 - x^2 = 0 \)
\( m = \frac{dy}{dx} = \frac{2xy \pm (x^2 + y^2)}{x^2 - y^2} \)
+ve \( \frac{dy}{dx} = \frac{(x + y)^2}{x^2 - y^2} = \frac{x + y}{x - y} \) ;
-ve \( \frac{dy}{dx} = -\frac{(x - y)^2}{x^2 - y^2} = -\frac{x - y}{x + y} \)
Homogenous \( y = tx \)
\( x \frac{dt}{dx} = \frac{1 + t^2}{1 - t} \)
\( \frac{1}{1 + t^2} dt - \frac{1}{2} \left( \frac{2t}{1 + t^2} \right) dt = \frac{dx}{x} \)
\( \tan^{-1} t - \frac{1}{2} \ln (1 + t^2) = \ln x + \ln k \)
\( \sqrt{x^2 + y^2} = c e^{\tan^{-1} y/x} \)
Similarly if we take -ve sign
\( \sqrt{x^2 + y^2} = c e^{-\tan^{-1} y/x} \)

Question. Let y = y(t) be a solution to the differential equation \( y' + 2t y = t^2 \), then find \( \lim_{t \to \infty} \frac{y}{t} \).
Answer:
\( y = y(t) \)
\( \frac{dy}{dt} + 2yt = t^2 \)
\( \text{I.F.} = e^{\int 2t dt} = e^{t^2} \)
\( y e^{t^2} = \int t^2 e^{t^2} dt \)
\( y = \frac{\int t^2 e^{t^2} dt}{e^{t^2}} \)
\( \lim_{t \to \infty} \frac{y}{t} = \lim_{t \to \infty} \frac{\int t^2 e^{t^2} dt}{t e^{t^2}} = \lim_{t \to \infty} \frac{t^2 e^{t^2}}{e^{t^2} + 2t^2 e^{t^2}} \)
\( = \lim_{t \to \infty} \frac{t^2}{2t^2 + 1} = \frac{1}{2} \)

Question. \( (1 - x^2) \frac{dy}{dx} + 2xy = x(1 - x^2)^{1/2} \)
Answer:
\( \frac{dy}{dx} + \left( \frac{2x}{1 - x^2} \right) y = \frac{x}{\sqrt{1 - x^2}} \)
\( \text{I.F.} = e^{\int \frac{2x}{1 - x^2} dx} = e^{-\ln(1 - x^2)} = \frac{1}{1 - x^2} \)
\( y \cdot \frac{1}{1 - x^2} = \int \frac{x}{(1 - x^2)^{3/2}} dx \)
\( \frac{y}{1 - x^2} = \frac{1}{\sqrt{1 - x^2}} + C \)
\( y = C(1 - x^2) + \sqrt{1 - x^2} \)

Question. Find the curve such that the area of the trapezium formed by the co-ordinate axes, ordinate of an arbitrary point and the tangent at this point equals half the square of its abscissa.
Answer:
Tangent
\( Y - y = \frac{dy}{dx} (X - x) \)
put \( X = 0 \)
\( \implies \) \( Y = \left( y - x \frac{dy}{dx} \right) \)
\( A = \frac{1}{2} x^2 \)
\( \frac{1}{2} \left[ 2y - x \frac{dy}{dx} \right] \cdot x = \frac{1}{2} x^2 \)
\( 2y - x \frac{dy}{dx} = \pm x \)
\( \frac{dy}{dx} - \frac{2}{x} y = \pm 1 \)
\( \text{IF} = e^{\int -\frac{2}{x} dx} = \frac{1}{x^2} \)
\( \left( \frac{y}{x^2} \right) = \int \pm \frac{1}{x^2} dx \)
\( \frac{y}{x^2} = \pm \left( -\frac{1}{x} \right) + c \)
\( y = cx^2 \pm x \)

Question. Find the curve possessing the property that the intercept, the tangent at any point of a curve cuts off on the y-axis is equal to the square of the abscissa of the point of tangency.
Answer:
\( Y - y = m (X - x) \)
\( X = 0, Y = y - mx \)
\( y - mx = x^2 \)
\( y - x \frac{dy}{dx} = x^2 \)
\( \frac{ydx - xdy}{x^2} = dx \)
\( d \left( \frac{y}{x} \right) = -dx \)
\( \frac{y}{x} = -x + C \)
\( y = cx - x^2 \)

Question. \( x \frac{dy}{dx} - y = 2x^2 \operatorname{cosec} 2x \)
Answer:
\( \frac{dy}{dx} - \frac{y}{x} = 2x \operatorname{cosec} 2x \)
\( \text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} \)
\( \frac{y}{x} = \int 2 \operatorname{cosec} 2x dx \)
\( \frac{y}{x} = \ln \tan x + C \)
\( y = cx + x \ln \tan x \)

Question. \( (1 + y^2) dx = (\tan^{-1} y - x) dy \)
Answer:
Put \( \tan^{-1} y = t \)
\( \implies \) \( \left( \frac{1}{1 + y^2} \right) dy = dt \)
\( dx = (t - x) dt \)
\( \implies \) \( \frac{dx}{dt} + x = t \)
\( \text{I.F.} = e^{\int 1 dt} = e^t \)
\( x e^t = \int t e^t dt \)
\( x e^t = t e^t - e^t + c \)
\( x = t - 1 + c e^{-t} \)
\( x = \tan^{-1} y - 1 + c e^{-\tan^{-1} y} \)

Question. Find the curve such that the area of the rectangle constructed on the abscissa of any point and the initial ordinate of the tangent at this point is equal to \( a^2 \). (Initial ordinate means y intercept of the tangent).
Answer:
y - Intercept of tangent \( = y - mx \)
\( x(y - mx) = \pm a^2 \)
\( xy - x^2 \frac{dy}{dx} = \pm a^2 \)
\( x^2 \frac{dy}{dx} = xy \mp a^2 \)
\( \frac{dy}{dx} = \frac{y}{x} \mp \frac{a^2}{x^2} \)
\( \frac{dy}{dx} - \frac{y}{x} = \mp \frac{a^2}{x^2} \)
\( \text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} \)
\( \frac{y}{x} = \pm \int \frac{a^2}{x^3} dx \)
\( \frac{y}{x} = \pm \left( -\frac{a^2}{2x^2} \right) + C \)
\( y = Cx \pm \frac{a^2}{2x} \)

Question. Let the function \( \ln f(x) \) is defined where \( f(x) \) exists for \( x \ge 2 \) and k is fixed positive real number, prove that if \( \frac{d}{dx} (x \cdot f(x)) \le - k f(x) \) then \( f(x) \le A x^{-1-k} \) where A is independent of x.
Answer:
\( \frac{d}{dx} (x f(x)) \le -k f(x) \)
\( x f'(x) + f(x) \le -k f(x) \)
\( x f'(x) + (1 + k) f(x) \le 0 \)
multiply by \( x^k \)
\( x^{k + 1} f'(x) + (k + 1) x^k f(x) \le 0 \)
\( \frac{d}{dx} (x^{k + 1} \cdot f(x)) \le 0 \)
Let \( F(x) = x^{k + 1} f(x) \)
F(x) is decreasing for \( x \ge 2 \)
\( F(x) \le F(2) \)
\( F(x) \le A \)
\( x^{k + 1} f(x) \le A \)
\( f(x) \le A \cdot x^{-1 - k} \)

Question. Find the differentiable function which satisfies the equation \( f(x) = -\int_{0}^{x} f(t) \tan t dt + \int_{0}^{x} \tan(t - x) dt \) where \( x \in (-\pi/2, \pi/2) \)
Answer:
put \( t - x = z \)
\( \implies \) \( dt = dz \)
\( f(x) = - \int_{0}^{x} f(t) \tan t dt + \int_{-x}^{0} \tan z dz \)
Now use leibnitz
\( f'(x) = -f(x) \tan x - \tan x \)
\( \frac{dy}{dx} + y \tan x = -\tan x \)
\( \text{I.F.} = e^{\int \tan x dx} = \sec x \)
\( y(\sec x) = - \int \tan x \sec x dx = -\sec x + c \)
\( y = c \cos x - 1 \)
\( y(0) = 0 \)
\( \implies \) \( c = 1 \)
\( y = \cos x - 1 \)

Question. Find all functions \( f(x) \) defined on \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \) with real values and has primitive \( F(x) \) such that \( f(x) + \cos x \cdot F(x) = \frac{\sin 2x}{(1 + \sin x)^2} \). Find \( f(x) \).
Answer:
Given that \( F(x) = \int f(x) dx \)
\( \implies \) \( f(x) = F'(x) \)
\( F'(x) + \cos x \cdot F(x) = \frac{\sin 2x}{(1 + \sin x)^2} \)
\( \frac{dy}{dx} + y \cos x = \frac{\sin 2x}{(1 + \sin x)^2} \)
\( \text{I.F.} = e^{\int \cos x dx} = e^{\sin x} \)
\( y(e^{\sin x}) = \int \frac{2 \sin x \cos x}{(1 + \sin x)^2} e^{\sin x} dx \)
put \( \sin x = t \)
\( = 2 \int \frac{t}{(1 + t)^2} e^t dt \)
\( y (e^{\sin x}) = \frac{2 e^{\sin x}}{1 + \sin x} + c \)
\( F(x) = \frac{2}{1 + \sin x} + c e^{-\sin x} \)
\( F'(x) = \frac{-2 \cos x}{(1 + \sin x)^2} - c e^{-\sin x} \cos x \)

Question. A tank contains 100 litres of fresh water. A solution containing 1 gm/litre of soluble lawn fertilizer runs into the tank at the rate of 1 lit/min and the mixture is pumped out of the tank of 3 litres/min. Find the time when the amount of fertilizer in the tank is maximum.
Answer:
Rate at which fertilizer added = \( 1 \times 1 = 1 \) gm/min.
volume of solution at time t = \( 100 + (1 - 3) t = 100 - 2t \)
\( \implies \) \( \frac{dy}{dt} = 1 - \left( \frac{3y}{100 - 2t} \right) \)
\( \frac{dy}{dt} + \left( \frac{3}{100 - 2t} \right) y = 1 \)
\( \implies \) \( y = (100 - 2t) + c (100 - 2t)^{3/2} \)
at \( t = 0, y = 0 \)
\( \implies \) \( c = -\frac{1}{10} \)
\( y = (100 - 2t) - \frac{1}{10} (100 - 2t)^{3/2} \)
\( \frac{dy}{dt} = 0 \)
\( \implies \) \( t = 27 \frac{7}{9} \)

Question. A tank with a capacity of 1000 litres originally contains 100 gms of salt dissolved in 400 litres of water. Beginning at time t = 0 and ending at time t = 100 minutes, water containing 1 gm of salt per litre enters the tank at the rate of 4 litre/minute and the wheel mixed solution is drained from the tank at a rate of 2 litre/minute. Find the differential equation for the amount of salt y in the tank at time t.
Answer:
Rate at which salt added = \( 4 \times 1 = 4 \) gm/min.
volume of solution at time t = \( 400 + (4 - 2)t = 400 + 2t \)
\( \frac{dy}{dt} = 4 - \left( \frac{y}{400 + 2t} \right) 2 \)
\( \frac{dy}{dt} = 4 - \frac{y}{200 + t} \)

Question. \( (x - y^2) dx + 2xy dy = 0 \)
Answer:
\( \frac{dy}{dx} = \frac{y}{2x} - \frac{1}{2y} \); put \( y^2 = t \)
\( 2y \frac{dy}{dx} = \frac{y^2}{x} - 1 \)
\( 2y \frac{dy}{dx} = \frac{dt}{dx} \)
\( \frac{dt}{dx} = \frac{t}{x} - 1 \)
\( \frac{dt}{dx} - \frac{t}{x} = -1 \)
\( \text{I.F.} = e^{\int -\frac{1}{x} dx} = \frac{1}{x} \)
\( \frac{t}{x} = -\int \frac{1}{x} dx \)
\( \frac{t}{x} = -\ln x - \ln C \)
\( \frac{y^2}{x} + \ln (xC) = 0 \)
\( y^2 + x \ln (xc) = 0 \)

Question. \( \frac{dy}{dx} - \frac{\tan y}{1 + x} = (1 + x) e^x \sec y \)
Answer:
\( \cos y \frac{dy}{dx} - \frac{\sin y}{1 + x} = (1 + x) e^x \)
\( \sin y = t \)
\( \implies \) \( \cos y \frac{dy}{dx} = \frac{dt}{dx} \)
\( \frac{dt}{dx} - \frac{t}{1 + x} = (1 + x) e^x \)
\( \text{I.F.} = e^{\int -\frac{1}{1 + x} dx} = \frac{1}{1 + x} \)
\( \frac{t}{1 + x} = \int e^x dx \)
\( \frac{t}{1 + x} = e^x + C \)
\( \frac{\sin y}{1 + x} = e^x + C \)
\( \sin y = (1 + x) (e^x + C) \)

Question. \( \frac{dy}{dx} = \frac{e^y}{x^2} - \frac{1}{x} \)
Answer:
\( \frac{1}{e^y} \frac{dy}{dx} = \frac{1}{x^2} - \frac{e^{-y}}{x} \)
\( e^{-y} = t \)
\( e^{-y} \frac{dy}{dx} = -\frac{dt}{dx} \)
\( -\frac{dt}{dx} = \frac{1}{x^2} - \frac{t}{x} \)
\( \frac{dt}{dx} - \frac{t}{x} = -\frac{1}{x^2} \)
\( \text{I.F.} = \frac{1}{x} \)
\( \implies \) \( \frac{t}{x} = -\int \frac{1}{x^3} dx \)
\( \frac{t}{x} = \frac{1}{2x^2} + C \)
\( 2x e^{-y} = 1 + Cx^2 \)

Question. \( \left( \frac{dy}{dx} \right)^2 - (x + y) \frac{dy}{dx} + xy = 0 \)
Answer:
\( \frac{dy}{dx} = \frac{(x + y) \pm \sqrt{(x + y)^2 - 4xy}}{2} \)
\( \frac{dy}{dx} = \frac{(x + y) \pm (x - y)}{2} \)
+ve \( \frac{dy}{dx} = x \)
\( \implies \) \( y = \frac{x^2}{2} + C \)
-ve \( \frac{dy}{dx} = y \)
\( \implies \) \( \ln y = x + C \)
\( y = k e^x \)

Question. \( \frac{dy}{dx} = \frac{y^2 - x}{2y(x + 1)} \)
Answer:
\( 2y \frac{dy}{dx} = \frac{y^2}{(x + 1)} - \frac{x}{(x + 1)} \)
Put \( y^2 = t \)
\( \implies \) \( 2y \frac{dy}{dx} = \frac{dt}{dx} \)
from \( \frac{dt}{dx} - \frac{t}{(x + 1)} = \frac{-x}{(x + 1)} \)
\( \text{I.F.} = \frac{1}{(x + 1)} \)
\( \frac{t}{(x + 1)} = -\int \frac{x}{(x + 1)^2} dx \)
\( = -\int \frac{dx}{x + 1} + \int \frac{dx}{(x + 1)^2} \)
\( \frac{y^2}{(x + 1)} = -\ln (x + 1) - \frac{1}{(x + 1)} + \ln C \)
\( y^2 = (x + 1) \ln \frac{C}{(x + 1)} - 1 \)

Question. \( \frac{dy}{dx} = e^{x - y} (e^x - e^y) \)
Answer:
\( \frac{dy}{dx} = \frac{e^{2x}}{e^y} - e^x \)
\( e^y \frac{dy}{dx} + e^x e^y = e^{2x} \)
\( e^y = t \)
\( \implies \) \( e^y \frac{dy}{dx} = \frac{dt}{dx} \)
\( \frac{dt}{dx} + e^x t = e^{2x} \)
\( \text{I.F.} = e^{\int e^x dx} = e^{e^x} \)
\( t e^{e^x} = \int e^{e^x} e^{2x} dx \)
\( \begin{matrix} e^x = z \\ e^x dx = dz \end{matrix} \)
\( = \int z e^z dz \)
\( t e^{e^x} = z e^z - e^z + C \)
\( t e^{e^x} = e^x e^{e^x} - e^{e^x} + C \)
\( e^y = (e^x - 1) + C \exp(-e^x) \)

Question. \( y y' \sin x = \cos x (\sin x - y^2) \)
Answer:
\( y \frac{dy}{dx} \sin x = \cos x \sin x - y^2 \cos x \)
\( y^2 = t \)
\( \implies \) \( y \frac{dy}{dx} = \frac{1}{2} \frac{dt}{dx} \)
\( \frac{1}{2} \frac{dt}{dx} \sin x = \cos x \sin x - t \cos x \)
\( \frac{dt}{dx} + 2t \cot x = 2 \cos x \)
\( \text{I.F.} = e^{\int 2 \cot x dx} = (\sin x)^2 \)
\( t (\sin x)^2 = 2 \int \cos x (\sin x)^2 dx \)
\( t (\sin x)^2 = \frac{2 (\sin x)^3}{3} + C \)
\( y^2 = \frac{2}{3} (\sin x) + \frac{C}{(\sin x)^2} \)

Advanced Subjective Questions

Question. \( \frac{dy}{dx} - y \ln 2 = 2^{\sin x} \cdot (\cos x - 1) \ln 2 \), \( y \) being bounded when \( x \to + \infty \).
Answer: I.f. = \( e^{\int \ln 2 dx} = (1/2)^x \)
\( \implies \) \( y(1/2)^x = \int 2^{(\sin x - x)} (\cos x - 1) \ln 2 dx \)
\( \implies \) let \( \sin x - x = t \)
\( \implies \) \( y \cdot \left( \frac{1}{2} \right)^x = 2^{\sin x - x} + c \)
\( \implies \) \( c = 0 \) as \( x \to \infty \) so \( y = 2^{\sin x} \)

Question. \( \frac{dy}{dx} = y + \int_0^1 y dx \) given \( y = 1 \), where \( x = 0 \)
Answer: Let \( A = \int_0^1 y dx \)
\( \implies \) \( \frac{dy}{dx} = y + A \)
\( \implies \) \( \frac{dy}{dx} - y = A \)
\( \implies \) If = \( e^{\int -1 dx} = e^{-x} \)
\( \implies \) \( y(e^{-x}) = A e^{-x} + c \)
\( \implies \) \( y = A + c e^x \)
\( \implies \) \( y = 1 \) when \( x = 0 \)
\( \implies \) \( c = 1 + A \)
\( \implies \) \( y = A + (1 + A) e^x \)
\( \implies \) \( A = \int_0^1 y dx \)
\( \implies \) \( A = \int_0^1 [A + (1 + A) e^x] dx \)
\( \implies \) \( A = \frac{e - 1}{3 - e} \)
\( \implies \) \( y = \frac{e - 1}{3 - e} + \left( 1 + \frac{e - 1}{3 - e} \right) e^x \)
\( \implies \) \( y = \frac{1}{3 - e} (2e^x - e + 1) \)

Question. Given two curves \( y = f(x) \) passing through the points \( (0, 1) \) & \( y = \int_{-\infty}^x f(t)dt \) passing through the points \( (0, 1/2) \). The tangents drawn to both curves at the points with equal abscissas intersect on the \( x \)-axis. Find the curve \( f(x) \).
Answer: Tangents at \( y = f(x) \) & \( y = \int_{-\infty}^x f(t) dt \)
\( \implies \) \( T_1 : y - f(x) = f'(x) (X - x) \)
\( \implies \) \( T_2 : Y - y_1 = f(x) (X - x) \) where \( \frac{dy_1}{dx} = f(x) \)
\( \implies \) from (i) & (ii) \( x \) coordinates are equal & \( y = 0 \)
\( \implies \) \( x - \frac{f(x)}{f'(x)} = x - \frac{y_1}{f(x)} \)
\( \implies \) \( \frac{f(x)}{f'(x)} = \frac{y_1}{f(x)} \)
\( \implies \) Integrating : \( \ln(y_1) = \ln(f(x) \cdot c) \)
\( \implies \) \( y_1 = f(x) \cdot c \)
\( \implies \) at \( x = 0 \), \( f(0) = 1 \) & \( y_1 = \int_{-\infty}^0 f(x) dx = 1/2 f(0) \)
\( \implies \) \( y_1 = 1/2 f(x) \)
\( \implies \) differentiating : \( f(x) = 1/2 f'(x) \)
\( \implies \) \( \frac{f'(x)}{f(x)} = 2 \)
\( \implies \) on integrating : \( f(x) = c_1 e^{2x} \)
\( \implies \) \( x = 0, f(0) = 1 \) so \( f(x) = e^{2x} \), \( c_1 = 1 \)

Question. Consider the differential equation, \( \frac{dy}{dx} + P(x)y = Q(x) \)
(i) If two particular solutions of given equation \( u(x) \) and \( v(x) \) are known, find the general solution of the same equation in terms of \( u(x) \) and \( v(x) \).
(ii) If \( \alpha \) and \( \beta \) are constants such that the linear combinations \( \alpha \cdot u(x) + \beta \cdot v(x) \) is a solution of the given equation, find the relation between \( \alpha \) and \( \beta \).
(iii) If \( w(x) \) is the third particular solution different from \( u(x) \) and \( v(x) \) then find the ratio \( \frac{v(x) - u(x)}{w(x) - u(x)} \).
Answer: (i) \( u' + Pu = Q \) and \( v' + Pv = Q \)
\( \implies \) subtract \( Q = \frac{uv' - vu'}{u - v} \)
\( \implies \) \( \int \frac{u' - v'}{u - v} = \int -P \)
\( \implies \) \( \ln(u - v) = - \int P dx \)
\( \implies \) \( u - v = e^{- \int P dx} \)
\( \implies \) \( IF = e^{\int P dx} = \frac{1}{u - v} \)
\( \implies \) \( y \left( \frac{1}{u - v} \right) = \int \frac{uv' - vu'}{(u - v)^2} dx \)
\( \implies \) \( y \left( \frac{1}{u - v} \right) = \frac{u}{u - v} + k \)
\( \implies \) \( y = u + k(u - v) \)
(ii) \( \alpha = 1 + k, \beta = -k \)
\( \implies \) \( \alpha + \beta = 1 \)
(iii) The ratio is a constant.

Question. \( x dy + y dx + \frac{x dy - y dx}{x^2 + y^2} = 0 \)
Answer: \( x dy + y dx = \frac{y dx - x dy}{x^2 + y^2} \)
\( \implies \) \( d(xy) = -d(\tan^{-1}(y/x)) \)
\( \implies \) \( xy + \tan^{-1}(y/x) = c \)

Question. \( \frac{y dx - x dy}{(x - y)^2} = \frac{dx}{2\sqrt{1 - x^2}} \), given that \( y = 2 \) when \( x = 1 \)
Answer: \( - \int d \left( \frac{y}{x - y} \right) = \int \frac{dx}{2\sqrt{1 - x^2}} \)
\( \implies \) \( - \left( \frac{y}{x - y} \right) = \frac{1}{2} \sin^{-1} x + c \)
\( \implies \) \( c = 2 - \pi/4 \) as \( y = 2 \) when \( x = 1 \)
\( \implies \) so \( \frac{\sin^{-1} x}{2} + \frac{y}{x - y} = \frac{\pi}{4} - 2 \)

Question. Find the continuous function which satisfies the relation, \( \int_0^x t f(x - t)dt = \int_0^x f(t)dt + \sin x + \cos x - x - 1 \), for all real number \( x \).
Answer: Let \( x - t = z \implies dt = -dz \)
\( \implies \) \( \int_0^x (x - z) f(z) dz = \int_0^x f(t) dt + \sin x + \cos x - x - 1 \)
\( \implies \) \( x \int_0^x f(z) dz - \int_0^x z f(z) dz = \int_0^x f(t) dt + \sin x + \cos x - x - 1 \)
\( \implies \) Use Leibnitz: \( \int_0^x f(z) dz + x f(x) - x f(x) = f(x) + \cos x - \sin x - 1 \)
\( \implies \) again Leibnitz: \( f(x) = f'(x) - \sin x - \cos x \)
\( \implies \) \( \frac{dy}{dx} - y = \sin x + \cos x \)
\( \implies \) \( IF = e^{-x} \)
\( \implies \) \( y(e^{-x}) = \int e^{-x} (\sin x + \cos x) dx \)
\( \implies \) \( y(e^{-x}) = -e^{-x} \cos x + c \)
\( \implies \) \( y = c e^x - \cos x \)
\( \implies \) \( x = 0 \implies y = 0 \implies c = 1 \)
\( \implies \) \( y = e^x - \cos x \)

Question. \( (1 - x^2)^2 dy + (y\sqrt{1 - x^2} - x - \sqrt{1 - x^2}) dx = 0 \).
Answer: \( \frac{dy}{dx} + y \frac{1}{(1 - x^2)^{3/2}} = \frac{x + \sqrt{1 - x^2}}{(1 - x^2)^{3/2}} \)
\( \implies \) I.F. = \( e^{\int \frac{1}{(1 - x^2)^{3/2}} dx} \)

Question. \( 3x^2 y^2 + \cos(xy) - xy \sin(xy) + \frac{dy}{dx} \{2x^3 y - x^2 \sin(xy)\} = 0 \).
Answer: \( 3x^2 y^2 dx + \cos(xy) dx - xy \sin(xy) dx + 2x^3 y dy - x^2 \sin(xy) dy = 0 \)
\( \implies \) \( d(x^3 y^2) + \cos xy dx - x \sin(xy) d(xy) = 0 \)
\( \implies \) \( d(x^3 y^2) + d(x \cdot \cos(xy)) = 0 \)
\( \implies \) integrate \( x^3 y^2 + x \cos(xy) = c \)
\( \implies \) \( x(x^2 y^2 + \cos(xy)) = c \)

Question. Find the integral curve of the differential equation, \( x(1 - x \ln y) \frac{dy}{dx} + y = 0 \) which passes through \( (1, 1/e) \).
Answer: \( y \frac{dx}{dy} + x = x^2 \log y \)
\( \implies \) dividing by \( x^2 y \)
\( \implies \) \( \frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = \frac{1}{y} \log y \)
\( \implies \) Let \( \frac{1}{x} = t \implies - \frac{1}{x^2} \frac{dx}{dy} = \frac{dt}{dy} \)
\( \implies \) \( - \frac{dt}{dy} + \frac{t}{y} = \frac{1}{y} \log y \)
\( \implies \) \( \frac{dt}{dy} - \frac{t}{y} = - \frac{1}{y} \log y \)
\( \implies \) If = \( e^{- \int \frac{1}{y} dy} = e^{-\log y} = 1/y \)
\( \implies \) \( t \cdot 1/y = \int - \frac{1}{y^2} \log y dy = \frac{1}{y} \log y - \int \frac{1}{y} \cdot \frac{1}{y} dy \)
\( \implies \) \( t = \log y + 1 + cy \implies \log ex + cy \)
\( \implies \) \( \frac{1}{x} = \log e y + cy \implies x(\log ey + cy) = 1 \)

Question. Find all the curves possessing the following property; the segment of the tangent between the point of tangency & the \( x \)-axis is bisected at the point of intersection with the \( y \)-axis.
Answer: Equation of tangent at \( (x, y) \) for \( y = f(x) \)
\( \implies \) \( Y - y = f'(x)(X - x) \).
\( \implies \) at \( x \)-axis point \( A = \left( -\frac{y}{f'(x)} + x, 0 \right) \)
\( \implies \) at \( y \)-axis point \( B = (0, -x f'(x) + y) \)
\( \implies \) given \( \frac{-\frac{y}{f'(x)} + x + x}{2} = 0 \implies -y + 2x f'(x) = 0 \)
\( \implies \) & \( \frac{0 + y}{2} = -x f'(x) + y \implies -2x f'(x) + y = 0 \)
\( \implies \) on solving the curve is \( y^2 = cx \)

Question. A perpendicular drawn from any point \( P \) of the curve on the \( x \)-axis meets the \( x \)-axis at \( A \). Length of the perpendicular from \( A \) on the tangent line at \( P \) is equal to 'a'. If this curve cuts the \( y \)-axis orthogonally, find the equation to all possible curves, expressing the answer explicitly.
Answer: Let the curve be \( y = f(x) \) & point \( P \) is \( (x, y) \)
\( \implies \) so point \( A \) is \( (x, 0) \)
\( \implies \) equation of tangent at \( P \) is \( Y - y = f'(x)(X - x) \)
\( \implies \) Length of \( \perp \) from \( (x, 0) \) to tangent is 'a'
\( \implies \) so \( \frac{y}{\sqrt{(f'(x))^2 + 1}} = a \)
\( \implies \) on squaring : \( y^2 = a^2 (f'(x))^2 + a^2 \)
\( \implies \) \( f'(x) = \pm \sqrt{\frac{y^2}{a^2} - 1} \)
\( \implies \) solve by taking +ve & -ve sign separately also on \( y \)-axis, \( x = 0 \)
\( \implies \) & angle b/w curve & \( y \)-axis is \( \frac{\pi}{2} \)
\( \implies \) so \( \implies f'(x) = 0 \)

Question. A curve passing through \( (1, 0) \) such that the ratio of the square of the intercept cut by any tangent off the \( y \)-axis to the subnormal is equal to the ratio of the product of the coordinates of the point of tangency to the product of square of the slope of the tangent and the subtangent at the same point. Determine all such possible curves.
Answer: Let the curve be \( y = f(x) \) & point of tangent is \( (x, y) \)
\( \implies \) Equation of tangent : \( Y - y = f'(x)(X - x) \)
\( \implies \) at \( y \)-axis, intercept \( = y - x f'(x) \)
\( \implies \) subnormal \( = y \cdot f'(x) \)
\( \implies \) Slope of tangent \( = f'(x) \)
\( \implies \) Subtangent \( = \frac{y}{f'(x)} \)
\( \implies \) according to question : \( \frac{(y - x f'(x))^2}{y \cdot f'(x)} = \frac{x \cdot y}{(f'(x))^2 \left( \frac{y}{f'(x)} \right)} \)
\( \implies \) Solve the equation & for \( C \), use point \( (1, 0) \)

Question. A & B are two separate reservoirs of water. Capacity of reservoir A is double the capacity of reservoir B. Both the reservoirs are filled completely with water, their inlets are closed and then the water is released simultaneously from both the reservoirs. The rate of flow of water out of each reservoir at any instant of time is proportional to the quantity of water in the reservoir at that time. One hour after the water is released, the quantity of water in reservoir A is 1.5 times the quantity of water in reservoir B. After how many hours do both the reservoirs have the same quantity of water ?
Answer: Let at any instant \( t \), \( x \) be the volume of water in reservoir A & \( y \) of that in B.
\( \implies \) \( \frac{dx}{dt} \propto x \implies \frac{dx}{dt} = k_1 x \implies x = e^{k_1 t} e^{C_1} \)
\( \implies \) Similarly \( \frac{dy}{dt} \propto y \implies y = e^{k_2 t} e^{C_2} \)
\( \implies \) Now at \( t = 0; x = 2y \implies x/y = 2 \)
\( \implies \) \( \frac{e^{C_1}}{e^{C_2}} = 2 \)
\( \implies \) also at \( t = 1; x = \frac{3}{2} y \implies \frac{x}{y} = \frac{3}{2} \)
\( \implies \) \( \frac{e^{k_1} e^{C_1}}{e^{k_2} e^{C_2}} = \frac{3}{2} \implies e^{k_1 - k_2} = 3/4 \)
\( \implies \) Let at \( t = T : x = y \implies \frac{x}{y} = 1 \) then \( \frac{e^{k_1 T} e^{C_1}}{e^{k_2 T} e^{C_2}} = 1 \)
\( \implies \) \( e^{(k_1 - k_2) T} = 1/2 \implies (3/4)^T = \frac{1}{2} \)
\( \implies \) \( T = \log_{4/3} 2 \)

Question. A tank consists of 50 litres of fresh water. Two litres of brine each litre containing 5 gms of dissolved salt are run into tank per minute; the mixture is kept uniform by stirring, and runs out at the rate of one litre per minute. If 'm' grams of salt are present in the tank after t minute, express 'm' in terms of t and find the amount of salt present after 10 minutes.
Answer: \( \frac{dm}{dt} = 10 - \left( \frac{m}{50 + t} \right) \)
\( \implies \) \( \frac{dm}{dt} + \frac{m}{50 + t} = 10 \)
\( \implies \) \( IF = e^{\int \frac{dt}{50 + t}} = 50 + t \)
\( \implies \) \( m(50 + t) = \int 10(50 + t)dt \)
\( \implies \) \( m(50 + t) = 10 \left( 50t + \frac{t^2}{2} \right) \)
\( \implies \) \( m = 5 \left( \frac{100t + t^2}{50 + t} \right) \)
\( \implies \) \( = 5t \left( \frac{100 + t}{50 + t} \right) \)
\( \implies \) \( = 5t \left( 1 + \frac{50}{50 + t} \right) \)
\( \implies \) at \( t = 10 \)
\( \implies \) \( m = 91 \frac{2}{3} \)

Question. Let \( f(x, y, c_1) = 0 \) and \( f(x, y, c_2) = 0 \) define two integral curves of homogeneous first order differential equation. If \( P_1 \) and \( P_2 \) are respectively the points of intersection of these curves with an arbitrary line, \( y = mx \) then prove that the slopes of these two curves at \( P_1 \) and \( P_2 \) are equal.
Answer: Assume two curves \( \frac{dy}{dx} = \frac{a_1 x + a_2 y}{a_3 x + a_4 y} \) & \( \frac{dy}{dx} = \frac{b_1 x + b_2 y}{b_3 x + b_4 y} \)
\( \implies \) Solve this differential equation by putting \( y = tx \) and then solve with \( y = mx \) to get \( P_1 \) and \( P_2 \) & get the slope at \( P_1 \) & \( P_2 \)

Question. Find the orthogonal trajectories for the given family of curves when 'a' is the parameter.
(i) \( y = ax^2 \)
(ii) \( \cos y = a e^{-x} \)
(iii) \( x^k + y^k = a^k \)
(iv) Find the isogonal trajectories for the family of rectangular hyperbolas \( x^2 - y^2 = a^2 \) which makes with it an angle of \( 45^\circ \).
Answer: (i) \( y = ax^2 \implies \frac{dy}{dx} = 2ax \)
\( \implies \) \( - \frac{dx}{dy} = 2ax \implies - \frac{dx}{dy} = \frac{2y}{x} \implies -x dx = 2y dy \)
\( \implies \) integrate \( \implies x^2 + 2y^2 = k \)
(ii) \( \cos y = a e^{-x} \)
\( \implies \) \( -\sin y \frac{dy}{dx} = -a e^{-x} \implies \sin y \frac{dy}{dx} = \cos y \implies \sin y \left( - \frac{dx}{dy} \right) = \cos y \)
\( \implies \) \( -dx = \cot y dy \implies \ln(\sin y) = -x + c \implies \sin y = k e^{-x} \)
(iii) \( x^k + y^k = a^k \implies kx^{k-1} + ky^{k-1} \cdot \frac{dy}{dx} = 0 \)
\( \implies \) \( \frac{dy}{dx} = - \frac{x^{k-1}}{y^{k-1}} \implies - \frac{dx}{dy} = - \frac{x^{k-1}}{y^{k-1}} \implies \frac{dx}{x^{k-1}} = \frac{dy}{y^{k-1}} \)
\( \implies \) After integrating \( \frac{1}{x^{k-2}} - \frac{1}{y^{k-2}} = \frac{1}{c^{k-2}} \) if \( k \neq 2 \)
(iv) \( x^2 - y^2 = a^2 \implies 2x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{y} \)
\( \implies \) \( \tan 45^\circ = \left| \frac{\frac{dy}{dx} - \frac{x}{y}}{1 + \frac{x}{y} \frac{dy}{dx}} \right| \)
\( \implies \) +ve : \( 1 + \frac{x}{y} \frac{dy}{dx} = \frac{dy}{dx} - \frac{x}{y} \)
\( \implies \) \( \frac{dy}{dx} = \frac{y + x}{y - x} \)
\( \implies \) after integration \( x^2 - y^2 + 2xy = c \)
\( \implies \) -ve : \( x^2 - y^2 - 2xy = c \)