Differential Equations JEE Mathematics Worksheets Set 01

Question. A solution of the differential equation, \( \left(\frac{dy}{dx}\right)^2 - x \frac{dy}{dx} + y = 0 \) is 
(a) y = 2
(b) y = 2x
(c) y = 2x - 4
(d) y = 2x^2 - 4
Answer: (c) y = 2x - 4
Solution:
\( \left(\frac{dy}{dx}\right)^2 - x \left(\frac{dy}{dx}\right) + y = 0 \)
\( \frac{dy}{dx} = \frac{x \pm \sqrt{x^2 - 4y}}{2} \)
or (c) option is statisfying given D.E.

Question. The differential equation representing the family of curves, \( y^2 = 2c(x + \sqrt{c}) \), where c is a positive parameter, is of
(a) order 1
(b) order 2
(c) degree 3
(d) degree 4
Answer: (c) degree 3
Solution:
\( y^2 = 2c(x + \sqrt{c}) \)
\( 2y \frac{dy}{dx} = 2c \)
\( \implies c = y \frac{dy}{dx} \)
\( y^2 = 2y \frac{dy}{dx} \left(x + \sqrt{y\frac{dy}{dx}}\right) \)
\( y^2 = 2yy'x + 2(yy')^{3/2} \)
\( (y^2 - 2yy'x)^2 = 4y^3(y')^3 \)
Order = 1 ; Degree = 3

Question. A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve.
Answer: Equation of Normal at point P (x, y) is
\( Y - y = -\frac{dx}{dy}(X - x) \) ....(1)
Perpendicular distance = |y|
\( \frac{\left|y + \frac{dx}{dy}.x\right|}{\sqrt{1 + \left(\frac{dx}{dy}\right)^2}} = |y| \)
\( y^2 + \left(\frac{dx}{dy}\right)^2 x^2 + 2xy \frac{dy}{dx} = y^2 \left(1 + \left(\frac{dx}{dy}\right)^2\right) \)
\( \left(\frac{dx}{dy}\right)^2 (x^2 - y^2) + 2xy \frac{dy}{dx} = 0 \)
\( \frac{dx}{dy} \left[ \frac{dx}{dy} (x^2 - y^2) + 2xy \right] = 0 \)
\( \frac{dx}{dy} = 0 \)
\( \implies x = c \)
x = 1
or \( \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \)
\( y = tx \)
\( \implies \frac{dy}{dx} = t + x \frac{dt}{dx} \)
\( t + x \frac{dt}{dx} = \frac{t^2 - 1}{2t} \)
\( x \frac{dt}{dx} = \frac{t^2 - 1 - 2t^2}{2t} \)
\( x \frac{dt}{dx} = \frac{-(t^2 + 1)}{2t} \)
\( \frac{2t}{t^2 + 1} dt = \frac{-dx}{x} \)
\( \ln (t^2 + 1) + \ln x = \ln C \)
\( x \frac{(y^2 + x^2)}{x^2} = C \)
(1, 1)
\( \implies C = 2 \)
\( y^2 + x^2 = 2x \)

Question. Solve the differential equation, \( (x^2 + 4y^2 + 4xy) dy = (2x + 4y + 1)dx \). 
Answer: \( \frac{dy}{dx} = \frac{2x + 4y + 1}{(x + 2y)^2} \)
\( \frac{dy}{dx} = \frac{2\left[x + 2y + \frac{1}{2}\right]}{(x + 2y)^2} \)
\( x + 2y = t \)
\( \implies 1 + 2 \frac{dy}{dx} = \frac{dt}{dx} \)
\( \frac{dt}{dx} - 1 = \frac{4t + 2}{t^2} \)
\( \frac{dt}{dx} = \frac{t^2 + 4t + 2}{t^2} \)
\( \frac{t^2}{(t + 2)^2 - 2} dt = dx \)
\( \frac{t^2}{(t + 2 + \sqrt{2})(t + 2 - \sqrt{2})} dt = dx \)
After Integrating
\( y = \ln [(x + 2y)^2 + 4(x + 2y) + 2] - \frac{3}{2\sqrt{2}} \ln \left( \frac{x + 2y + 2 - \sqrt{2}}{x + 2y + 2 + \sqrt{2}} \right) + C \)

Question. A country has a food deficit of 10%. Its population grows continuously at a rate of 3%. Its annual food production every year is 4% more than that of the last year. Assuming that the average food requirement per person remains constant, prove that the country will become self-sufficient in food after 'n' years, where 'n' is the smallest integer bigger than or equal to, \( \frac{\ln 10 - \ln 9}{\ln(1.04) - 0.03} \).
Answer: Let \( x_0 \) be initial population
\( y_0 \) be its initial food
Let avg. consumption be a unit.
Than food Reqd initially = \( ax_0 \)
\( y_0 = ax_0 (0.90) \)
\( y_0 = 0.9a X_0 \) ....(1)
Let x be the population of country in 'n' years.
\( \frac{dx}{dn} = \text{rate of change of population} = \frac{3}{100} x = 0.03 x \)
\( \frac{dx}{x} = 0.03 dn \)
\( \ln x = 0.03 dn + C \)
\( x = A e^{0.03n} \)
At \( n = 0, x = x_0 \)
\( \implies A = x_0 \)
\( x = x_0 e^{0.03n} \)
'y' be the food production in year t
\( y = y_0 \left(1 + \frac{4}{100}\right)^n = 0.9 a x_0 (1.04)^n \)
\( y_0 = 0.9 a x_0 \)
Food consumption in the year n is \( a x_0 e^{0.03n} \)
Again \( y - x \ge 0 \)
\( 0.9 x_0 a (1.04)^n > a x_0 e^{0.03n} \)
\( \frac{(1.04)^n}{e^{0.03n}} \ge \frac{1}{0.9} = \frac{10}{9} \)
\( n[\log (1.04) - 0.03] \ge \log 10 - \log 9 \)
\( n \ge \frac{\log 10 - \log 9}{\log(1.04) - \log 0.03} \)

Question. A hemispherical tank of radius 2 metres is initially full of water and has an outlet of \( 12\text{ cm}^2 \) cross sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law \( V(t) = 0.6 \sqrt{2gh(t)} \), where V(t) and h(t) are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time t, and g is the acceleration due to gravity. Find the time it takes to empty the tank. 
Answer: volume of water layer
\( dV = \pi(r^2 - h^2) dh \)
flux \( Q(t) = v(t) \times A \)
\( \implies (72/10^5) \sqrt{2}\sqrt{g}\sqrt{2 - h} \int dt = \pi \int \frac{2^2 - h^2}{\sqrt{2 - h}} dh \)
\( \implies (72/10^5) \sqrt{2}\sqrt{g}\sqrt{2 - h}.t + c \)
\( = -2\pi \left[ \frac{4}{3}(2 - h)^{3/2} - \frac{(2 - h)^{5/2}}{5} \right] \)
at t = 0 : h = 0
\( \implies c = \frac{-56\pi\sqrt{2}}{15} \)
so \( t = \frac{-56\pi\sqrt{2}}{15} \times \frac{10^5}{72 \times \sqrt{2}\sqrt{g}} = \frac{7\pi \times 10^5}{135\sqrt{g}}\text{ sec.} \)

Question. Find the equation of the curve which passes through the origin and the tangent to which at every point (x, y) has slope equal to \( \frac{x^4 + 2xy - 1}{1 + x^2} \). 
Answer: \( \frac{dy}{dx} = \frac{x^4 - 1 + 2xy}{1 + x^2} \)
\( \frac{dy}{dx} = (x^2 - 1) + \frac{2x}{1 + x^2} y \)
\( \frac{dy}{dx} - \frac{2x}{1 + x^2} y = x^2 - 1 \)
I.F. = \( e^{-\int \frac{2x}{1 + x^2} dx} = e^{-\ln(1 + x^2)} = \frac{1}{1 + x^2} \)
\( y \frac{1}{1 + x^2} = \int \frac{x^2 + 1 - 2}{x^2 + 1} dx \)
\( y \frac{1}{1 + x^2} = \int \frac{x^2 + 1 - 2}{x^2 + 1} dx = \int \left( 1 - \frac{2}{1 + x^2} \right) dx \)
\( \frac{y}{1 + x^2} = x - 2 \tan^{-1} x + c \)
passing through origin
\( \implies c = 0 \)
\( y = (x - 2 \tan^{-1} x) (1 + x^2) \)

Question. Let f(x), \( x \ge 0 \), be a nonnegative continuous function, and let \( F(x) = \int_{0}^{x} f(t) dt \), \( x \ge 0 \). If for some c > 0, \( f(x) \le cF(x) \) for all \( x \ge 0 \), then show that f(x) = 0 for all \( x \ge 0 \). 
Answer: \( f(x) \ge 0 \) and \( F(x) \ge 0 \) ...(1)
and also \( F'(x) = f(x) \) ...(2)
also \( f(x) \le cF(x) \)
\( \implies F'(x) \le cF(x) \) using (2)
\( \int_{0}^{x} \frac{F'(x)}{F(x)} \le \int_{0}^{x} c \)
\( \implies F(x) \le e^{cx} \) using (1) and (2)
\( F(x) = 0 \) :for all \( x \ge 0 \)
Aliter
\( F'(x) \le cF(x) \)
\( e^{-cx} dy/dx - c.e^{-cx} y \le 0 \)
\( \implies d/dx (e^{-cx} y) \le 0 \)
Let \( h(x) = F(x) . e^{-cx} \downarrow \)
\( \implies h(x) \le h(0) \)
\( \implies h(x) \le 0 \)
\( \implies F(x) \le 0 \)

Question. A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Find the time after which the cone is empty.
Answer: \( \frac{dv}{dt} \propto -s \)
\( \frac{dv}{dt} = -ks \)
volume of cone = \( \frac{1}{3} \pi r^2 h \)
surface area = \( \pi r^2 \)
\( \tan \theta = \frac{R}{H} = \frac{r}{h} \)
\( v = \frac{1}{3} \pi r^3 \cot \theta \) and \( s = \pi r^2 \)
\( \frac{1}{3} \cot \theta \pi 3r^2 \frac{dr}{dt} = -k \pi r^2 \)
\( \cot \theta \int_{R}^{0} dr = -k \int_{0}^{T} dt \)
\( R \cot \theta = KT \)
\( \implies H = KT \)
\( \implies T = H/K \)

Question. If P(1) = 0 and \( \frac{dP(x)}{dx} > P(x) \) for all \( x \ge 1 \) then prove that P(x) > 0 for all x > 1.
Answer: \( dy/dx - y > 0 \) (given)
multiply both sides by \( e^{-x} \)
\( e^{-x} dy/dx - e^{-x} y > 0 \)
\( \implies d/dx (e^{-x} y) > 0 \)
Let \( h(x) = P(x) . e^{-x} \uparrow \) : h(1) = 0
\( \implies h(x) > h(1) \)
\( \implies h(x) > 0 \)
\( \implies P(x) > 0 \)

Question. If \( \left(\frac{2 + \sin x}{1 + y}\right) \frac{dy}{dx} = -\cos x, y(0) = 1 \), then \( y\left(\frac{\pi}{2}\right) \) equals 
(a) 1
(b) 1/2
(c) 1/3
(d) 1/4
Answer: (c) 1/3
Solution:
\( \left(\frac{2 + \sin x}{1 + y}\right) \frac{dy}{dx} = -\cos x \)
\( \frac{dy}{1 + y} = -\frac{\cos x}{2 + \sin x} dx \)
\( \ln (1 + y) = -\log(2 + \sin x) + \ln c \)
\( x = 0, y = 1 \)
\( \implies c = 4 \)
\( \ln (1 + y) = -\ln 3 + \ln 4 \)
\( 1 + y = \frac{4}{3} \)
\( y = \frac{1}{3} \)

Question. A curve passes through (2, 0) and the slope of tangent at point P (x, y) equals \( \frac{(x + 1)^2 + y - 3}{(x + 1)} \). Find the equation of the curve and area enclosed by the curve and the x-axis in the fourth quadrant. 
Answer: \( \frac{dy}{dx} = \frac{(x + 1)^2 + y - 3}{(x + 1)} \)
\( \frac{dy}{dx} - \frac{y}{(x + 1)} = \frac{(x + 1)^2 - 3}{(x + 1)} \)
I.F. = \( e^{-\int \frac{dx}{x + 1}} = \frac{1}{x + 1} \)
\( \frac{y}{(x + 1)} = \int \frac{(x + 1)^2 - 3}{(x + 1)^2} dx \)
\( = \int \left( 1 - \frac{3}{(x + 1)^2} \right) dx \)
\( \frac{y}{(x + 1)} = x + \frac{3}{(x + 1)} + c \)
(2, 0) passes c = -3
\( y = x(x + 1) + 3 - 3(x + 1) \)
\( y = x^2 - 2x \)
Area = \( \left| \int_{0}^{2} (x^2 - 2x) dx \right| = 4/3\text{ sq. units} \)

Question. The solution of primitive integral equation \( (x^2 + y^2)dy = xy dx \), is y = y(x). If y(1) = 1 and \( y(x_0) = e \), then \( x_0 \) is
(a) \( \sqrt{2(e^2 - 1)} \)
(b) \( \sqrt{2(e^2 + 1)} \)
(c) \( \sqrt{3} e \)
(d) \( \sqrt{\frac{e^2 + 1}{2}} \)
Answer: (c) \( \sqrt{3} e \)
Solution:
\( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \)
\( y = tx \)
\( \implies \frac{dy}{dx} = t + x \frac{dt}{dx} \)
\( t + x \frac{dt}{dx} = \frac{t}{1 + t^2} \)
\( x \frac{dt}{dx} = \frac{t - t - t^3}{1 + t^2} \)
\( \frac{1 + t^2}{t^3} dt = -\frac{dx}{x} \)
Altar solving and passing through (1, 1)
\( -\frac{x^2}{2y^2} + \ln y = -\frac{1}{2} \)
\( y = e, x = x_0 \)
\( -\frac{x_0^2}{2e^2} + 1 = -\frac{1}{2} \)
\( x_0^2 = 3e^2 \)
\( \implies x_0 = \sqrt{3} e \)

Question. For the primitive integral equation ydx + y^2dy = xdy; x ∈ R, y > 0, y = y(x), y(1) = 1, then y(-3) is 
(a) 3
(b) 2
(c) 1
(d) 5
Answer: (a) 3
Solution:
\( ydx + y^2dy = xdy \)
\( ydx - xdy = -y^2dy \)
\( \frac{ydx - xdy}{y^2} = -dy \)
\( d\left(\frac{x}{y}\right) = -dy \)
\( \frac{x}{y} = -y + c \quad (1, 1) \)
\( \implies c = 2 \)
\( y^2 - 2y + x = 0 \)
\( y(-3) \)
\( y^2 - 2y - 3 = 0 \)
\( \implies y = 3, -1 \)

Question. If length of tangent at any point on the curve y = f(x) intercepted between the point and the x-axis is of length 1. Find the equation of the curve.
Answer: Length of tangent = \( \left| y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \right| = 1 \)
\( \implies y^2 \left( 1 + \left(\frac{dx}{dy}\right)^2 \right) = 1 \)
\( \frac{dy}{dx} = \pm \frac{y}{\sqrt{1 - y^2}} \)
\( \int \frac{\sqrt{1 - y^2}}{y} dy = \pm x + c \)
\( \sqrt{1 - y^2} + \ln \left| \frac{1 - \sqrt{1 - y^2}}{y} \right| = \pm x + c \)

Question. A tangent drawn to the curve, y = f(x) at P(x, y) cuts the x-axis and y-axis at A and B respectively such that BP : AP = 3 : 1, given that f(1) = 1, then
(a) equation of the curve is \( x\frac{dy}{dx} - 3y = 0 \)
(b) equation of curve is \( x\frac{dy}{dx} + 3y = 0 \)
(c) curve passes through (2, 1/8)
(d) normal at (1, 1) is x + 3y = 4
Answer: (b) equation of curve is \( x\frac{dy}{dx} + 3y = 0 \), (c) curve passes through (2, 1/8)
Solution:
Equation of tangent \( Y - y = \frac{dy}{dx}(X - x) \)
\( \frac{BP}{AP} = \frac{3}{1} \)
\( B(0, y - mx) \)
\( A(x - \frac{y}{m}, 0) \)
\( \frac{BP}{AP} = \frac{3}{1} \)
so, \( \frac{dx}{x} = -\frac{dy}{3y} \)
\( \ln x = -\frac{1}{3} \ln y - \ln c \)
\( \ln x^3 = -\ln cy \)
\( \frac{1}{x^3} = cy \quad f(1) = 1 \)
\( \implies c = 1 \)
\( y = \frac{1}{x^3} \)

Question. Let f(x) be differentiable on the interval \( (0, \infty) \) such that f(1) = 1 and \( \lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1 \) for each x > 0. Then f(x) is 
(a) \( \frac{1}{3x} + \frac{2x^2}{3} \)
(b) \( \frac{-1}{3x} + \frac{4x^2}{3} \)
(c) \( \frac{-1}{x} + \frac{2}{x^2} \)
(d) \( \frac{1}{x} \)
Answer: (a) \( \frac{1}{3x} + \frac{2x^2}{3} \)
Solution:
\( \lim_{t \to x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1 \)
\( x^2 f'(x) - 2x f(x) + 1 = 0 \)
\( \implies f(x) = cx^2 + \frac{1}{3x} \) \quad f(1) = 1
\( c = \frac{2}{3} \)
\( f(x) = \frac{2}{3} x^2 + \frac{1}{3x} \)

Question. The differential equation \( \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{y} \) determines a family of circles with
(a) variable radii and a fixed centre at (0, 1)
(b) variable radii and a fixed centre at (0, -1)
(c) fixed radius 1 and variable centres along the x-axis.
(d) fixed radius 1 and variable centres along the y-axis.
Answer: (c) fixed radius 1 and variable centres along the x-axis.
Solution:
\( \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{y} \)
\( \int \frac{y}{\sqrt{1 - y^2}} dy = \int dx \)
\( -\sqrt{1 - y^2} = x + c \)
\( (x + c)^2 + y^2 = 1 \)
centre (-c, 0) radius = \( \sqrt{c^2 - c^2 + 1} = 1 \)

Question. Let a solution y = y (x) of the differential equation, \( x\sqrt{x^2 - 1} dy - y \sqrt{y^2 - 1} dx = 0 \) satisfy \( y(2) = \frac{2}{\sqrt{3}} \).
STATEMENT-1 : \( y(x) = \sec \left(\sec^{-1} x - \frac{\pi}{6}\right) \)
and
STATEMENT-2 : y(x) is given by \( \frac{1}{y} = \frac{2\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^2}} \)
(a) Statement-1 is true, Statement-2 is true ; Statement-2 is correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true ; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Answer: (c) Statement-1 is true, Statement-2 is false.
Solution:
\( x\sqrt{x^2 - 1} dy - y \sqrt{y^2 - 1} dx = 0 \)
\( \frac{dx}{x\sqrt{x^2 - 1}} = \frac{dy}{y\sqrt{y^2 - 1}} \)
\( \sec^{-1} x = \sec^{-1} y + c \)
\( \sec^{-1} 2 = \sec^{-1} \frac{2}{\sqrt{3}} + c \)
\( c = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \)
\( \sec^{-1} x = \sec^{-1} y + \frac{\pi}{6} \)
\( y = \sec \left(\sec^{-1} x - \frac{\pi}{6}\right) \)
\( \cos^{-1} \frac{1}{x} = \cos^{-1} \frac{1}{y} + \frac{\pi}{6} \)
\( \cos^{-1} \frac{1}{y} = \cos^{-1} \frac{1}{x} - \cos^{-1} \left(\frac{\sqrt{3}}{2}\right) \)
\( \frac{1}{y} = \frac{\sqrt{3}}{2x} - \sqrt{1 - \frac{1}{x^2}} \left(\frac{1}{2}\right) \)
\( \implies \frac{2}{y} = \frac{\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^2}} \)

Question. If y(x) satisfies the differential equation \( y' - y \tan x = 2x \sec x \) and y(0) = 0, then 
(a) \( y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{8\sqrt{2}} \)
(b) \( y'\left(\frac{\pi}{4}\right) = \frac{\pi^2}{18} \)
(c) \( y\left(\frac{\pi}{3}\right) = \frac{\pi^2}{9} \)
(d) \( y'\left(\frac{\pi}{3}\right) = \frac{4\pi}{3} + \frac{2\pi^2}{3\sqrt{3}} \)
Answer: (a) \( y\left(\frac{\pi}{4}\right) = \frac{\pi^2}{8\sqrt{2}} \), (d) \( y'\left(\frac{\pi}{3}\right) = \frac{4\pi}{3} + \frac{2\pi^2}{3\sqrt{3}} \)
Solution:
I.F. = \( \cos x \)
\( y . \cos x = \int 2x \sec x . \cos x . dx \)
\( y . \cos x = x^2 + c \), c = 0
\( y = x^2 \sec x \)

Question. State the order & degree of the following differential equations :
(i) \( \left[ \frac{d^2x}{dt^2} \right]^3 + \left[ \frac{dx}{dt} \right]^4 - xt = 0 \)
(ii) \( \frac{d^2y}{dx^2} = \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{3/2} \)
Answer: (i) \( \left[ \frac{d^2x}{dt^2} \right]^3 + \left[ \frac{dx}{dt} \right]^4 - xt = 0 \)
Order = 2
Degree = 3
(ii) \( \frac{d^2y}{dx^2} = \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{3/2} \)
\( \implies \) \( \left( \frac{d^2y}{dx^2} \right)^2 = \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^3 \)
Order = 2
Degree = 3

Question. \( \frac{\ln(\sec x + \tan x)}{\cos x} dx = \frac{\ln(\sec y + \tan y)}{\cos y} dy \)
Answer: \( \frac{\ln(\sec x + \tan x)}{\cos x} dx = \frac{\ln(\sec y + \tan y)}{\cos y} dy \)
\( \int \sec x \ln(\sec x + \tan x) dx = \int \sec y \ln(\sec y + \tan y) dy \)
\( [\ln(\sec x + \tan x)]^2 = [\ln(\sec y + \tan y)]^2 + k \)

Question. \( \frac{dy}{dx} + \frac{\sqrt{(x^2 - 1)(y^2 - 1)}}{xy} = 0 \)
Answer: \( \frac{dy}{dx} + \frac{\sqrt{(x^2 - 1)(y^2 - 1)}}{xy} = 0 \)
\( \frac{y dy}{\sqrt{y^2 - 1}} + \frac{\sqrt{x^2 - 1}}{x} dx = 0 \)
\( \frac{y dy}{\sqrt{y^2 - 1}} + \frac{x^2 - 1}{x\sqrt{x^2 - 1}} dx = 0 \)
\( \frac{y dy}{\sqrt{y^2 - 1}} + \frac{x}{\sqrt{x^2 - 1}} dx - \frac{1}{x\sqrt{x^2 - 1}} dx = 0 \)
Integrate
\( \sqrt{y^2 - 1} + \sqrt{x^2 - 1} - \sec^{-1} x = C \)

Question. \( \frac{dy}{dx} = \sin (x + y) + \cos (x + y) \)
Answer: \( \frac{dy}{dx} = \sin (x + y) + \cos (x + y) \)
Put \( x + y = t \)
\( 1 + \frac{dy}{dx} = \frac{dt}{dx} \)
\( \frac{dt}{dx} - 1 = \sin t + \cos t \)
\( \frac{dt}{dx} = \sin t + \cos t + 1 \)
\( \frac{dt}{\sin t + \cos t + 1} = dx \)
\( \frac{\sec^2 \frac{t}{2} dt}{1 + \tan \frac{t}{2}} = 2dx \)
\( \ln \left( 1 + \tan \frac{t}{2} \right) = x + C \)
\( \ln \left( 1 + \tan \frac{x + y}{2} \right) = x + C \)

Question. (a) \( \frac{dy}{dx} + \sin \frac{x + y}{2} = \sin \frac{x - y}{2} \)
(b) \( \sin x \cdot \frac{dy}{dx} = y \cdot \ln y \) if \( y = e \), when \( x = \frac{\pi}{2} \)
Answer: (a) \( \frac{dy}{dx} + \sin \frac{x + y}{2} = \sin \frac{x - y}{2} \)
\( \frac{dy}{dx} = - 2 \cos \frac{x}{2} \sin \frac{y}{2} \)
\( \int \operatorname{cosec} \frac{y}{2} dy = - \int 2 \cos \frac{x}{2} dx \)
\( \ln \tan \frac{y}{4} = C - 2 \sin \frac{x}{2} \)

(b) \( \sin x \frac{dy}{dx} = y \ln y \)
\( \frac{dy}{y \ln y} = \operatorname{cosec} x dx \)
\( \ln (\ln y) = \ln \left( \tan \frac{x}{2} \right) + C \)
\( \left( \frac{\pi}{2}, e \right) \)
\( \implies \) \( C = 0 \)
\( \ln y = \tan \frac{x}{2} \)
\( \implies \) \( y = e^{\tan \frac{x}{2}} \) Particular solution.

Question. \( e^{(dy/dx)} = x + 1 \) given that when \( x = 0 \), \( y = 3 \)
Answer: \( e^{dy/dx} = x + 1 \)
\( \frac{dy}{dx} = \ln (x + 1) \)
\( \int dy = \int \ln(x + 1) dx \)
\( y = (x + 1) [\ln (x + 1) - 1] + C \)
\( (0, 3) \)
\( \implies \) \( C = 4 \)
\( y = (x + 1) \ln (x + 1) - x - 1 + 4 \)
\( y = (x + 1) \ln (x + 1) - x + 3 \) Particular solution.

Question. The population P of a town decreases at a rate proportional to the number by which the population exceeds 1000, proportionality constant being k > 0. Find
(a) Population at any time t, given initial population of the town being 2500.
(b) If 10 years later the population has fallen to 1900, find the time when the population will be 1500.
(c) Predict about the population of the town in the long run.
Answer: (a) \( \frac{-dP}{dt} = k (P - 1000) \)
\( \ln (P - 1000) = -kt + c \)
\( P = 1000 + c_1 e^{-kt} \)
\( t = 0 \)
\( \implies \) \( P = 2500 \)
\( \implies \) \( c_1 = 1500 \)
\( P = 1000 + 1500 e^{-kt} \)

(b) \( t = 10 \)
\( \implies \) \( P = 1900 \)
\( 900 = 1500 e^{-k(10)} \)
\( \implies \) \( k = \frac{1}{10} \ln \left( \frac{5}{3} \right) \)
\( \implies \) \( P = 1000 + 1500 e^{-kt} \)
\( 1500 = 1000 + 1500 e^{-kt} \)
\( \implies \) \( e^{kt} = 3 \)
\( \implies \) \( kt = \ln 3 \)
\( t = \frac{\ln 3}{k} = 10 \frac{\ln 3}{\ln \left( \frac{5}{3} \right)} \)
\( = 10 \log_{5/3} 3 \)

Question. It is known that the decay rate of radium is directly proportional to its quantity at each given instant. Find the law of variation of a mass of radium as a function of time if at t = 0, the mass of the radius was \( m_0 \) and during time \( t_0 \) \( \alpha\% \) of the original mass of radium decay.
Answer: \( \frac{-dm}{dt} = Km \)
\( \ln m = - Kt + \ln C \)
\( m = e^{-Kt} \cdot e^{\ln C} \)
\( \implies \) \( m = C e^{-Kt} \)
at \( t = 0, m = m_0 \)
\( \implies \) \( C = m_0 \)
\( m = m_0 e^{-Kt} \)
at \( t = t_0, m = m_0 \left( 1 - \frac{\alpha}{100} \right) \)
\( m_0 \left( 1 - \frac{\alpha}{100} \right) = m_0 e^{-K t_0} \)
\( \implies \) \( K = \frac{-1}{t_0} \ln \left( 1 - \frac{\alpha}{100} \right) \)
\( m = m_0 e^{-Kt} \)
Where
\( K = \frac{-1}{t_0} \ln \left( 1 - \frac{\alpha}{100} \right) \)

Question. A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k, then show that the differential equation describing such curves is, \( y \frac{dy}{dx} = \pm \sqrt{k^2 - y^2} \). Find the equation of such a curve passing through (0, k).
Answer: \( L_N \) (Length of the normal) \( = y \sqrt{1 + m^2} \)
\( y \sqrt{1 + m^2} = k \)
\( \implies \) \( 1 + m^2 = \frac{k^2}{y^2} \)
\( m^2 = \frac{k^2 - y^2}{y^2} \)
\( \implies \) \( m = \pm \frac{\sqrt{k^2 - y^2}}{y} \)
\( y \frac{dy}{dx} = \pm \sqrt{k^2 - y^2} \)
\( \implies \) \( \frac{y dy}{\sqrt{k^2 - y^2}} = \pm dx \)
\( -\sqrt{k^2 - y^2} = \pm x + C \)
(0, k)
\( \implies \) \( C = 0 \)
\( -\sqrt{k^2 - y^2} = \pm x \)
\( k^2 - y^2 = x^2 \)
\( x^2 + y^2 = k^2 \) Circle

Question. Find the curve for which the sum of the lengths of the tangent and subtangent at any of its point is proportional to the product of the co-ordinates of the point of tangency, the proportionality factor is equal to k.
Answer: \( \frac{y \sqrt{1 + m^2}}{m} + \frac{y}{m} = kxy \)
\( \frac{\sqrt{1 + m^2} + 1}{m} = kx \)
\( \sqrt{1 + m^2} = (kxm - 1) \)
\( 1 + m^2 = k^2x^2m^2 + 1 - 2kxm \)
\( m = 0 \) ; \( m = \frac{-2kx}{1 - k^2x^2} \)
\( \implies \) \( \frac{dy}{dx} = \frac{-2kx}{1 - k^2x^2} \)
\( dy = \frac{1}{k} \left[ \frac{-2k^2x}{1 - k^2x^2} \right] dx \)
\( y = \frac{1}{k} \ln |(1 - k^2x^2)| + \ln C \)

Question. Find the curve y = f(x) where f(x) ≥ 0, f(0) = 0, bounding a curvilinear trapezoid with the base [0, x] whose area is proportional to (n + 1)th power of f(x). It is known that f(1) = 1.
Answer: \( A = \int_{0}^{x} f(x) dx = k [f(x)]^{n+1} \)
Use Leibnitz
\( f(x) = k (n + 1) [f(x)]^n \cdot f'(x) \)
\( \frac{1}{k(n + 1)} = [f(x)]^{n-1} \cdot f'(x) \)
\( \frac{x}{k(n + 1)} = \frac{[f(x)]^n}{n} + C \)
\( x = 0, f(0) = 0 \)
\( \implies \) \( C = 0 \)
\( x = 1, f(1) = 1 \)
\( \implies \) \( k = \frac{1}{n + 1} \)
\( [f(x)]^n = x \)
\( f(x) = x^{1/n} \)

Question. A curve is such that the length of the polar radius of any point on the curve is equal to the length of the tangent drawn at this point. Form the differential equation and solve it to find the equation of the curve.
Answer: Given that
\( y \frac{\sqrt{1 + m^2}}{m} = \sqrt{x^2 + y^2} \)
\( y^2 (1 + m^2) = m^2 (x^2 + y^2) \)
\( y^2 = m^2x^2 \)
\( m = \pm \frac{y}{x} \)
\( \frac{dy}{dx} = \frac{y}{x} \); \( \frac{dy}{dx} = \frac{-y}{x} \)
\( \frac{dy}{y} = \frac{dx}{x} \); \( \frac{dy}{y} = -\frac{dx}{x} \)
\( \ln y = \ln x + \ln k \); \( \ln y = -\ln x + \ln k \)
\( \frac{y}{x} = k' \)
\( \implies \) \( y = k'x \)
\( xy = k' \)

Question. (a) \( \frac{dy}{dx} = \frac{x^2 + xy}{x^2 + y^2} \)
(b) \( (x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy \)
Answer: (a) \( \frac{dy}{dx} = \frac{x^2 + xy}{x^2 + y^2} \)
\( \implies \) \( \frac{dy}{dx} = \frac{1 + y/x}{1 + (y/x)^2} \)
Put \( y = tx \)
\( \implies \) \( \frac{dy}{dx} = t + x \frac{dt}{dx} \)
\( t + x \frac{dt}{dx} = \frac{1 + t}{1 + t^2} \)
\( \implies \) \( x \frac{dt}{dx} = \frac{1 + t}{1 + t^2} - t \)
\( x \frac{dt}{dx} = \frac{1 + t - t - t^3}{1 + t^2} \)
\( \implies \) \( \frac{1 + t^2}{1 - t^3} dt = \frac{dx}{x} \)
\( \frac{t^2}{1 - t^3} dt + \frac{1}{1 - t^3} dt = \frac{dx}{x} \)
\( \frac{t^2}{1 - t^3} dt + \frac{1 - t^2 + t^2}{1 - t^3} dt = \frac{dx}{x} \)
\( \frac{2t^2}{1 - t^3} dt + \frac{(1 - t)(1 + t)}{(1 - t)(1 + t + t^2)} dt = \frac{dx}{x} \)
\( -\frac{2}{3} \int \frac{-3t^2}{1 - t^3} dt + \frac{1}{2} \int \frac{2t + 1}{t^2 + t + 1} dt + \frac{1}{2} \int \frac{1}{t^2 + t + 1} dt = \frac{dx}{x} \)
Integrate both the side
\( -\frac{2}{3} \ln (1 - t^3) + \frac{1}{2} \ln (t^2 + t + 1) + \frac{1}{2 \cdot \frac{\sqrt{3}}{2}} \tan^{-1} \left( \frac{2t + 1}{\sqrt{3}} \right) = \ln x + C \)
After putting value of 't'
\( C (x - y)^{2/3} (x^2 + xy + y^2)^{1/6} = \exp \left[ \frac{1}{\sqrt{3}} \tan^{-1} \frac{x + 2y}{x\sqrt{3}} \right] \)
(b) \( (x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy \)
\( \frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} \)
\( \frac{dy}{dx} = \frac{1 - 3(y/x)^2}{(y/x)^3 - 3(y/x)} \)
Put \( y = tx \)
\( \implies \) \( \frac{dy}{dx} = t + x \frac{dt}{dx} \)
\( t + x \frac{dt}{dx} = \frac{1 - 3t^2}{t^3 - 3t} \)
\( x \frac{dt}{dx} = \frac{1 - 3t^2 - t^4 + 3t^2}{t(t^2 - 3)} \)
\( \frac{t(t^2 - 3)}{1 - t^4} dt = \frac{dx}{x} \)
After integration
\( y^2 - x^2 = c (y^2 + x^2)^2 \)

Question. Find the equation of a curve such that the projection of its ordinate upon the normal is equal to its abscissa.
Answer: \( x = \left| y \frac{1}{\sqrt{1 + m^2}} \right| \)
\( \frac{y}{\sqrt{1 + m^2}} = \pm x \)
\( \implies \) \( 1 + m^2 = \frac{y^2}{x^2} \)
\( m^2 = \frac{y^2 - x^2}{x^2} \)
\( \implies \) \( \frac{dy}{dx} = \pm \frac{\sqrt{y^2 - x^2}}{x} \)
Put \( y = t x \)
\( \implies \) \( \frac{dy}{dx} = t + x \frac{dt}{dx} \)
\( \frac{dt}{\pm \sqrt{t^2 - 1} - t} = \frac{dx}{x} \)
After integration
\( \frac{y^2 \pm y\sqrt{y^2 - x^2}}{x^2} = \ln \left[ \frac{(y \pm \sqrt{y^2 - x^2}) c^2}{x^3} \right] \)

Question. The perpendicular from the origin to the tangent at any point on a curve is equal to the abscissa of the point of contact. Find the equation of the curve satisfying the above condition and which passes through (1, 1).
Answer: \( Y - y = m (X - x) \)
\( Y - y - m (X - x) = 0 \)
\( \left| \frac{-y + mx}{\sqrt{1 + m^2}} \right| = x \)
\( (mx - y)^2 = x^2 (1 + m^2) \)
\( m^2x^2 - 2mxy + y^2 = x^2 + x^2m^2 \)
\( 2mxy = y^2 - x^2 \)
\( m = \frac{y^2 - x^2}{2xy} \)
\( \frac{dy}{dx} = \frac{\frac{y^2}{x^2} - 1}{\frac{2y}{x}} \)
Put \( y = tx \)
\( t + x \frac{dt}{dx} = \frac{t^2 - 1}{2t} \)
\( x \frac{dt}{dx} = \frac{t^2 - 1 - 2t^2}{2t} \)
\( \frac{2t}{-t^2 - 1} dt = \frac{dx}{x} \)
\( \ln (t^2 + 1) = -\ln x + \ln C \)
\( x (t^2 + 1) = C \)
\( x \left( \frac{y^2 + x^2}{x^2} \right) = C \)
\( C = 2 \) (1, 1) Passes
\( y^2 + x^2 = 2x \)
\( x^2 + y^2 - 2x = 0 \)

Question. Show that the equation of the curve intersecting with the x-axis at the point x = 1 and for which the length of the subnormal at any point of the curve is equal to the arithmetic mean of the co-ordinates of this point is (y - x)^2 (x + 2y) = 1.
Answer: \( |y m| = \frac{x + y}{2} \)
+ve \( y \frac{dy}{dx} = \frac{x + y}{2} \)
\( 2 \frac{dy}{dx} = \frac{x + y}{y} \)
Put \( y = tx \)
\( 2 \left[ t + x \frac{dt}{dx} \right] = \frac{1 + t}{t} \)
\( 2t + 2x \frac{dt}{dx} = \frac{1 + t}{t} \)
\( \implies \) \( 2x \frac{dt}{dx} = \frac{1 + t - 2t^2}{t} \)
\( \frac{2t}{1 + t - 2t^2} dt = \frac{dx}{x} \)
After integrating and then passes through (1, 0) will get
\( (y - x)^2 (x + 2y) = 1 \)

Question. Use the substitution \( y^2 = a - x \) to reduce the equation \( y^3 \cdot \frac{dy}{dx} + x + y^2 = 0 \) to homogeneous form and hence solve it.
Answer: \( y^2 = a - x \)
\( 2y \frac{dy}{dx} = \frac{da}{dx} - 1 \)
\( y^3 \frac{dy}{dx} + x + y^2 = 0 \)
\( y^2 \left( 2y \frac{dy}{dx} \right) + 2x + 2y^2 = 0 \)
\( (a - x) \left( \frac{da}{dx} - 1 \right) + 2x + 2(a - x) = 0 \)
\( (a - x) \left( \frac{da}{dx} - 1 \right) + 2a = 0 \)
\( \frac{da}{dx} - 1 = \frac{-2a}{a - x} \)
\( \implies \) \( \frac{da}{dx} = \frac{-a - x}{a - x} \)
Put \( a = tx \)
\( \frac{da}{dx} = t + x \frac{dt}{dx} \)
\( t + x \frac{dt}{dx} = \frac{-1 - a/x}{a/x - 1} \)
\( x \frac{dt}{dx} = \frac{-1 - t}{t - 1} - t \)
\( x \frac{dt}{dx} = \frac{-1 - t - t^2 + t}{t - 1} \)
\( \frac{t - 1}{-1 - t^2} dt = \frac{dx}{x} \)
\( \implies \) \( \frac{1 - t}{1 + t^2} dt = \frac{dx}{x} \)
\( \left( \frac{1}{1 + t^2} \right) dt - \frac{1}{2} \left( \frac{2t}{1 + t^2} \right) dt = \frac{dx}{x} \)
\( \tan^{-1} t - \frac{1}{2} \ln (1 + t^2) = \ln x + \ln C \)
\( \tan^{-1} \frac{a}{x} - \frac{1}{2} \ln \left( \frac{a^2 + x^2}{x^2} \right) = \ln x + \ln C \)
\( \tan^{-1} \frac{a}{x} - \frac{1}{2} \ln (a^2 + x^2) = \ln C \)
\( \frac{1}{2} \ln (a^2 + x^2) - \tan^{-1} \frac{a}{x} = k \)
Where \( a = y^2 + x \)

Question. \( \left[ x \cos \frac{y}{x} + y \sin \frac{y}{x} \right] y = \left[ y \sin \frac{y}{x} - x \cos \frac{y}{x} \right] x \frac{dy}{dx} \)
Answer: \( \left[ x \cos \frac{y}{x} + y \sin \frac{y}{x} \right] y = \left[ y \sin \frac{y}{x} - x \cos \frac{y}{x} \right] x \frac{dy}{dx} \)
\( \frac{dy}{dx} = \frac{\left[ x \cos \frac{y}{x} + y \sin \frac{y}{x} \right] y}{\left[ y \sin \frac{y}{x} - x \cos \frac{y}{x} \right] x} \)
\( \frac{dy}{dx} = \frac{\left[ \cos \frac{y}{x} + \frac{y}{x} \sin \frac{y}{x} \right] \frac{y}{x}}{\left[ \frac{y}{x} \sin \frac{y}{x} - \cos \frac{y}{x} \right]} \)
Put \( y = tx \)
\( x \frac{dt}{dx} = \frac{t[\cos t + t \sin t]}{t \sin t - \cos t} - t \)
\( x \frac{dt}{dx} = \frac{t \cos t + t^2 \sin t - t^2 \sin t + t \cos t}{t \sin t - \cos t} \)
\( \frac{t \sin t - \cos t}{2t \cos t} dt = \frac{dx}{x} \)
\( -\ln (t \cos t) = 2 \ln x + \ln C \)
\( \ln (t \cos t) x^2 = \ln k \)
\( (t \cos t) x^2 = k' \)
\( \left( \frac{y}{x} \right) \cos \left( \frac{y}{x} \right) x^2 = k' \)
\( \implies \) \( xy \cos \left( \frac{y}{x} \right) = k' \)

Question. \( (x - y) dy = (x + y + 1) dx \)
Answer: \( \frac{dy}{dx} = \frac{(x + y + 1)}{x - y} \)
Let \( x = X + h \)
\( \implies \) \( dx = dX \)
\( y = Y + k \)
\( \implies \) \( dy = dY \)
\( \frac{dY}{dX} = \frac{(X + h) + (Y + k) + 1}{(X + h) - (Y + k)} \)
\( = \frac{(X + Y) + (h + k + 1)}{(X - Y) + (h - k)} \)
\( h + k + 1 = 0 \) & \( h - k = 0 \)
\( \implies \) \( h = -\frac{1}{2}, k = -\frac{1}{2} \)
\( \frac{dY}{dX} = \frac{X + Y}{X - Y} \)
Put \( Y = tX \)
\( X \frac{dt}{dX} = \frac{1 + t}{1 - t} - t \)
\( \implies \) \( \frac{1 - t}{1 + t^2} dt = \frac{dX}{X} \)
\( \tan^{-1} t - \frac{1}{2} \ln (1 + t^2) = \ln X + \ln C \)
\( \tan^{-1} t - \frac{1}{2} \ln (Y^2 + X^2) = \ln C \)
\( \tan^{-1} \frac{Y}{X} = \ln C \sqrt{Y^2 + X^2} \)
\( \tan^{-1} \left( \frac{y + 1/2}{x + 1/2} \right) = \ln C \sqrt{\left( x + \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2} \)
\( \tan^{-1} \left( \frac{2y + 1}{2x + 1} \right) = \ln C \sqrt{x^2 + y^2 + x + y + \frac{1}{2}} \)

Question. \( \frac{dy}{dx} = \frac{x + y + 1}{2x + 2y + 3} \)
Answer: \( \frac{dy}{dx} = \frac{x + y + 1}{2x + 2y + 3} \)
\( \frac{dy}{dx} = \frac{x + y + 1}{2(x + y) + 3} \)
Put \( x + y + 1 = t \)
\( 2 \left[ \frac{dt}{dx} - 1 \right] = \frac{t}{t + 1/2} \); \( 1 + \frac{dy}{dx} = \frac{dt}{dx} \)
\( \frac{dt}{dx} - 1 = \frac{t}{2t + 1} \)
\( \frac{dt}{dx} = \frac{3t + 1}{2t + 1} \)
\( \left( \frac{2t + 1}{3t + 1} \right) dt = dx \)
\( \frac{2}{3} \left[ \frac{3t + 3/2}{3t + 1} \right] dt = dx \)
\( \frac{2}{3} \left[ 1 + \frac{1}{2(3t + 1)} \right] dt = dx \)
\( \frac{2}{3} t + \frac{1}{9} \ln (3t + 1) = x + C \)
\( (3t + 1) = k e^{3(x - 2y)} \)
\( 3(x + y + 1) + 1 = k e^{3(x - 2y)} \)
\( x + y + \frac{4}{3} = k' e^{3(x - 2y)} \)