Samacheer Class 12 Maths Volume 2 - Ordinary Differential Equations

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EXERCISE 10.1EXERCISE 10.11 questions

Q.1For each of the following differential equations, determine its order and degree (if exists): (i) dy/dx + x y = cot x. (ii) (d^3 y/dx^3)^2 − (d^2 y/dx^2)^3 + 5(dy/dx)^4 = 0. (iii) (contains sin of derivatives — non‑polynomial). (iv) (highest derivative order 7, appears linearly). (v) (highest derivative order 3, appears linearly). (vi) (equation linear in highest derivative of order 2). (vii) (highest derivative order 2 appears linearly). (viii) (contains cos of derivative — non‑polynomial). (ix) (contains an integral — non‑algebraic in derivatives). (x) x e^{x} y (dy/dx) = ... (derivative appears polynomially).v

Solution

Concise reasoning: - Order = highest order derivative occurring. - Degree = power of the highest order derivative after clearing fractions and making the equation a polynomial in derivatives; if the equation involves non‑algebraic functions of derivatives (sin, cos, integrals, exponentials multiplying derivatives in nonpolynomial way) degree is not defined. Apply these rules to each part: (i) has dy/dx linearly → order 1, degree 1. (ii) highest derivative is d^3y/dx^3 appearing squared → order 3 degree 2. (iii) and (viii),(ix) involve nonpolynomial functions → degree does not exist. For the remaining parts the highest derivative orders are as stated above and appear to first power → degree 1.

Answer:

(i) order = 1, degree = 1. (ii) order = 3, degree = 2. (iii) order = 2, degree does not exist (non‑polynomial because of sin). (iv) order = 7, degree = 1 (highest derivative assumed 7 and appears linearly). (v) order = 3, degree = 1. (vi) order = 2, degree = 1. (vii) order = 2, degree = 1. (viii) order = 2, degree does not exist (contains cos — non‑polynomial in derivatives). (ix) order = 1, degree does not exist (contains an integral — not algebraic in derivatives). (x) order = 1, degree = 1 (derivative appears to first power).

EXERCISE 10.2EXERCISE 10.22 questions

Q.1Express each physical statement as a differential equation: (i) Radium decays at a rate proportional to the amount Q present. (ii) Population P increases at a rate proportional to P(500000−P). (iii) For a substance, rate of change of vapour pressure P wrt temperature T is proportional to P and inversely proportional to T^2. (iv) A sum grows at 8% per year (continuous compounding) and receives continuously an additional ₹400 per year.v

Solution

Direct translation of proportionality statements: rate = constant × indicated quantity (with negative sign for decay). For (iv) continuous 8% gives 0.08A; continuous credit ₹400 adds +400 to dA/dt.

Answer:

(i) \(\dfrac{dQ}{dt}=-kQ\), \(k>0\). (ii) \(\dfrac{dP}{dt}=kP(500000-P)\). (iii) \(\dfrac{dP}{dT}=k\dfrac{P}{T^2}\). (iv) If A(t) is amount, \(\dfrac{dA}{dt}=0.08A+400\).

Q.2Assume a spherical raindrop evaporates at a rate proportional to its surface area. Form a differential equation for the rate of change of its radius r(t).v

Solution

Differentiate V: dV/dt = 4π r^2 (dr/dt). Equate to -k·S = -k·4π r^2 and cancel 4π r^2 (r>0) to get dr/dt = -k (constant).

Answer:

Let V=(4/3)π r^3 and surface area S=4π r^2. dV/dt = -k S = -4π k r^2 ⇒ 4π r^2 dr/dt = -4π k r^2 ⇒ dr/dt = -k.

EXERCISE 10.3EXERCISE 10.38 questions

Q.1Find the differential equation of the family of (i) all non‑vertical lines in a plane (ii) all non‑horizontal lines in a plane.v

Solution

A straight line has two arbitrary constants; eliminating them by differentiation twice gives the second derivative zero. For lines expressed as x(y) the same idea gives d^2x/dy^2=0.

Answer:

(i) For non‑vertical lines y=mx+c ⇒ y''=0. (ii) For non‑horizontal lines x=ny+c (view x as function of y) ⇒ d^2x/dy^2=0. (If expressed as y(x) and excluding horizontal lines, the family is still y=mx+c with m≠0 and y''=0.)

Q.2Form the differential equation of all straight lines touching the circle x^2+y^2=r^2.v

Solution

Tangent condition: distance from centre to line equals radius; algebraically for y=mx+c tangency gives c^2=r^2(1+m^2). Replace m by y' and c by y-xy' (value of c from y=mx+c) to obtain the DE shown.

Answer:

For line y=mx+c to be tangent to circle: c^2=r^2(1+m^2). Using m=y' and c=y-xy' gives differential equation: (y-xy')^2 = r^2(1+(y')^2).

Q.3Find the differential equation of the family of circles passing through the origin and having their centres on the x‑axis.v

Solution

General circle centre (a,0) passing through (0,0): (x-a)^2+y^2=a^2 ⇒ x^2+y^2-2ax=0. Differentiate: 2x+2yy'-2a=0 ⇒ a=x+yy'. Substitute a into original: x^2+y^2-2x(x+yy')=0 ⇒ y^2-x^2-2xyy'=0 ⇒ y^2-x^2=2xyy'.

Answer:

Differential equation: y^2 - x^2 = 2 x y y' .

Q.4Find the differential equation of the family of all the parabolas with latus rectum 4a and whose axes are parallel to the x‑axis.v

Solution

General parabola with axis parallel to x‑axis: (y-k)^2=4a(x-h). Differentiate: 2(y-k)y'=4a ⇒ y-k=2a/y'. Differentiate again: 2(y')^2+2(y-k)y''=0. Substitute y-k from above: 2(y')^2+2(2a/y')y''=0 ⇒ (y')^3+2a y''=0.

Answer:

Differential equation: (y')^3 + 2a y'' = 0.

Q.5Find the differential equation of the family of parabolas with vertex at (0,-1) and having axis along the y‑axis.v

Solution

Parabola with axis along y: x^2=4a(y+1). Differentiate: 2x = 4a y' ⇒ a = x/(2y'). Substitute into original: x^2 = 4(x/(2y'))(y+1) ⇒ x y' = 2(y+1) ⇒ x y' - 2y - 2 = 0.

Answer:

Differential equation: x y' = 2y + 2.

Q.6Find the differential equations of the family of all ellipses having foci on the y‑axis and centre at the origin.v

Solution

General ellipse: x^2/a^2 + y^2/b^2 =1. Differentiate: x/a^2 + y y'/b^2 =0 ⇒ express 1/b^2 and substitute into original to obtain b^2 = y^2 - x y y'. Since b is constant differentiate b^2: d/dx(y^2 - x y y')=0 ⇒ 2y y' - [y y' + x((y')^2 + y y'')] =0. Simplify to x y y'' + x (y')^2 - y y' =0.

Answer:

Differential equation: x y y'' + x (y')^2 - y y' = 0.

Q.7Find the differential equation corresponding to the family of curves represented by y = A e^{x/8} + B e^{-x/8}, where A and B are arbitrary constants.v

Solution

Differentiate twice: y'' = (1/64)(A e^{x/8} + B e^{-x/8}) = (1/64) y ⇒ 64 y'' - y = 0.

Answer:

Differential equation: 64 y'' - y = 0 (equivalently y'' - (1/64) y = 0).

Q.8Find the differential equation of the curve represented by x y = a e^{x} + b e^{-x} + ... (OCR ambiguous: interpret as xy = a e^{x} + b e^{-x} + ...).v

Solution

Standard two‑parameter exponential family y=Ae^x+Be^{-x} satisfies y''=Ae^x+Be^{-x}=y, hence y''-y=0. If the OCR intended a different form, please provide the exact expression for a precise DE.

Answer:

If the family is y = A e^{x} + B e^{-x} then the DE is y'' - y = 0. (For xy = a e^{x}+b e^{-x} more information needed; if interpreted as y = A e^x + B e^{-x} the DE is y''-y=0.)

EXERCISE 10.4EXERCISE 10.48 questions

Q.1Show that each given expression is a solution of the differential equation. (i) y = x^2 for x y' = 2y. (ii) y = A e^{x} + B e^{-x} for y'' − y = 0.v

Solution

Differentiate the proposed solutions and substitute into the given differential equations to verify identity.

Answer:

(i) y=x^2 ⇒ y'=2x ⇒ x y' = x·2x = 2x^2 = 2y, holds. (ii) y=Ae^x+Be^{-x} ⇒ y''=Ae^x+Be^{-x}=y ⇒ y''-y=0, holds.

Q.2Find value(s) of m so that y = e^{m x} is a solution of: (i) y' + y = 0. (ii) y'' − y' − 6y = 0.v

Solution

Substitute y=e^{mx} and cancel e^{mx} (nonzero) to obtain algebraic equations for m and solve.

Answer:

(i) Interpreting as y'+y=0 ⇒ for y=e^{mx}: m e^{mx}+e^{mx}=0 ⇒ m+1=0 ⇒ m=-1. (ii) For y=e^{mx}: m^2 - m - 6 = 0 ⇒ (m-3)(m+2)=0 ⇒ m=3 or m=-2.

Q.3The slope of the tangent is the reciprocal of four times the ordinate: dy/dx = 1/(4y). The curve passes through (2,5). Find the equation of the curve.v

Solution

Separable ODE solved by integration and constant determined from initial point.

Answer:

Integrate: dy/dx = 1/(4y) ⇒ 4y dy = dx ⇒ ∫4y dy = ∫ dx ⇒ 2 y^2 = x + C. Use (2,5): 2·25 = 2 + C ⇒ 50 = 2 + C ⇒ C = 48. Hence 2 y^2 = x + 48 or x = 2 y^2 - 48.

Q.4Show that y = e^{mx} + n x - 1 (OCR interpreted) is a solution of the differential equation e^x d y/dx (x^2 - 1) = 0 (?) (OCR ambiguous).v

Solution

Unable to verify because the OCR text of the DE is unclear. Provide precise equation and the verification will be done by differentiating y and substituting.

Answer:

OCR is ambiguous; typical exercise: show y = A e^{mx} + n x + c satisfies a linear ODE by substitution. Please provide exact differential equation for verification.

Q.5Show that y = a x + b (a ≠ 0) is a solution of x y'' + x y' − y = 0.v

Solution

Differentiate y and substitute; constant term forces b=0. If the exercise had a different DE, please supply exact form.

Answer:

Substitute y=ax+b ⇒ y'=a, y''=0. Then x·0 + x·a - (ax+b) = xa - ax - b = -b. For this to be 0, b=0. Thus general linear y=ax (b must be 0). If the intended DE was x y'' + x y' - y = 0, y = ax is a solution; y=ax+b is solution only if b=0.

Q.6Show that y = A e^{bx} + C (OCR: y = a e^{bx} + b?) where a,b are arbitrary constants, is a solution of d y/dx + (dy/dx)^2 = 3 ? (OCR ambiguous).v

Solution

Cannot proceed without the precise differential equation; provide correct statement for verification.

Answer:

Given OCR ambiguity, unable to confirm. If the intended DE is dy/dx = b a e^{bx} etc. Please provide exact DE.

Q.7Show that the differential equation representing the family y = a sqrt(x) / (1+ a^2) (OCR ambiguous) is y - x y y' ... (given RHS).v

Solution

Provide clearer OCR of the family; then eliminate the parameter by differentiating once and substituting to obtain the required differential equation.

Answer:

OCR ambiguous. From the textbook the family y = a x /(1+a^2) (?) leads to a specific algebraic DE after eliminating parameter a. Please provide the exact family expression to derive the DE.

Q.8Show that y = a cos(bx) is a solution of the differential equation d^2y/dx^2 + b^2 y = 0.v

Differentiate: y' = -ab sin(bx), y'' = -ab^2 cos(bx) = -b^2(a cos(bx)) = -b^2 y.

Hence y'' + b^2 y = 0, so y = a cos(bx) satisfies the equation.

EXERCISE 10.5EXERCISE 10.54 questions

Q.1If a constant force F acts on a car of mass M with resisting force proportional to velocity kV, equation M dV/dt = F - kV. Find V(t) given V(0)=0.v

Solution

Linear ODE: dV/dt + (k/M)V = F/M. Integrating factor e^{(k/M)t}. Solve: V e^{(k/M)t} = (F/k)(e^{(k/M)t}-1) ⇒ V=(F/k)(1-e^{-(k/M)t}). Initial condition V(0)=0 satisfied.

Answer:

Solution: V(t) = (F/k) (1 - e^{- (k/M) t}).

Q.2The velocity v of a parachute falling vertically satisfies the equation v (dv/dx) = g - (k/2) v^2, where g and k are constants. If v and x are both initially zero, find v in terms of x.v

Solution

Given v dv/dx = g - (k/2) v^2. Put A = k/2. Then v/(g - A v^2) dv = dx. Integrate: \int v/(g-A v^2) dv = -\frac{1}{2A}\ln(g-A v^2)=x+C. With v(0)=0: -\frac{1}{2A}\ln g = C. Hence \ln(g-A v^2)=\ln g -2Ax, so g-A v^2 = g e^{-2Ax}. Thus v^2 = (g/A)(1-e^{-2Ax}). Substituting A=k/2 gives v^2=(2g/k)(1-e^{-kx}). Taking positive root (downward speed): v=\sqrt{\dfrac{2g}{k}(1-e^{-kx})}.

Answer:

v = \sqrt{\dfrac{2g}{k}\bigl(1-e^{-kx}\bigr)}

Q.3Find the equation of the curve whose slope is (y-x)/(x+2) and which passes through the point (1,0).v

Solution

We interpret dy/dx = (y-x)/(x+2). Rewrite: dy/dx - y/(x+2) = -x/(x+2). IF = e^{-\int dx/(x+2)}=1/(x+2). Multiply: d/dx(y/(x+2)) = -x/(x+2)^2. Integrate: y/(x+2) = \int -\frac{x}{(x+2)^2}dx = -\ln(x+2) -\frac{2}{x+2} + C. Hence y=-(x+2)\ln(x+2) -2 + C(x+2). Use (1,0): 0= -3\ln3 -2 +3C => C=(2+3\ln3)/3. Substitute for final y.

Answer:

y = -(x+2)\ln(x+2) -2 + C(x+2) with C=\tfrac{2+3\ln 3}{3}; so y=-(x+2)\ln(x+2)-2 + (\tfrac{2+3\ln 3}{3})(x+2).

Q.4Solve the following differential equations: (i) dy dx y x = − − 2 (ii) ydx x x dy++() = − 1 0 2 1 tan (iii) sin, dy dx a y = () = 0 1 (iv) dy dx e x e x y y =++3 (v) e x dx e x dy y y +()+= 1 0 cos sin (vi) ydx xdy x y ny dx − ()       = cot 2 (vii) dy dx x x − − = 25 0 2 (viii) x y dy e x x dx x cos log =+() 1 (ix) tan cos cos y dy dx x y x y =+() +− () (x) dy dx x y =+() tan 2v

Solution

The printed OCR for parts (i)–(x) is too garbled to uniquely reconstruct the intended differential equations. Please resend the ten subparts with clear formatting (for example: (i) dy/dx = ..., (ii) ...). Once the exact equations are available I will provide concise solutions.

Answer:

Cannot solve — question text is ambiguous due to OCR errors. Please provide clear statements for each subpart (i)–(x).

EXERCISE 10.6EXERCISE 10.65 questions

Q.1Exactness / integrating factor problem (OCR too garbled). Original text unreadable.v

Solution

The given line contains mixed symbols and is not unambiguous. To proceed I need the differential equation in a clear form (e.g. M(x,y)dx + N(x,y)dy = 0). Please re-submit the equation.

Answer:

Cannot solve — OCR of the equation is unclear. Please provide the exact differential equation in plain text.

Q.3Equation beginning with ye dx + xe y dy ... (OCR ambiguous).v

Solution

OCR has scrambled the terms. Provide the differential equation in standard form (M dx + N dy = 0) so I can check exactness or compute an integrating factor and solve.

Answer:

Cannot solve — equation not unambiguously readable. Please retype the differential equation.

Q.5Two differential equations presented (OCR unclear).v

Solution

The OCR text is insufficiently clear to reconstruct the intended problems. If you paste the exact equations (for example: (a) M(x,y)dx + N(x,y)dy = 0 ; (b) ...), I'll solve them concisely.

Answer:

Cannot solve — please provide clear statement(s) of the differential equation(s).

Q.7Equation involving exponentials: '1 3 3 1 0+ ( ) + - ( ) = e dy e y x dx y x y x, given that y=0 when x=1' (OCR ambiguous).v

Solution

Unable to parse the coefficients and terms from the OCR. Provide the equation in the form M(x,y)dx + N(x,y)dy = 0 (or dy/dx = ...), and specify the initial condition, then I will give the solution.

Answer:

Cannot solve — equation text ambiguous. Please re-enter the differential equation clearly.

Q.8x y dy + x y dx = ? It is given that y(1)=1 and y(x0)=e; find x0. (OCR ambiguous)v

Solution

The posted equation is unclear. Provide the differential equation explicitly (e.g. x^2 y dy + x y dx = 0 or similar) and the two conditions; I will compute x0.

Answer:

Cannot determine — equation ambiguous. Please give the exact differential equation.

EXERCISE 10.7EXERCISE 10.79 questions

Q.1cos x + sin x dy/dx = ? The OCR is jumbled: 'cos sin x dy dx y x+= 1 2. 1 1 2 − () − = x dy dx xy'.v

Solution

The OCR merges two separate lines; send each differential equation in clear form (e.g. dy/dx + P(x)y = Q(x)) and I will solve them.

Answer:

Cannot solve — please provide a clear single differential equation.

Q.3dy/dx = y + sin x (OCR: 'dy dx y x x+= sin 4. x dy dx xy x 2 2 1 2 4+()+=+')v

Solution

The OCR text appears to combine multiple problems. Provide the intended differential equation(s) unambiguously.

Answer:

Cannot solve — unclear; please restate each equation separately and clearly.

Q.5Two equations: '2 10 0 3 x y dy ydx − ()+=' and 'x x dy dx x x x y x sin cos sin sin' (OCR unclear).v

Solution

The OCR is unreadable. Provide the exact differential equations in plain text and I will produce concise solutions.

Answer:

Cannot solve — please supply clear equations.

Q.7y e^x dy/dx ... (OCR unclear): 'y e dx dy x x − ()+− = − sin 1 1 0 2'.v

Solution

Provide the ODE as M(x,y)dx + N(x,y)dy = 0 or dy/dx = ..., with clear functions, and I will solve it.

Answer:

Cannot solve — equation ambiguous. Please retype cleanly.

Q.91 0 2 3 ++()++() = x xy dy dx y y (OCR unclear).v

Solution

OCR corrupted; please type the differential equation exactly.

Answer:

Cannot solve — please give clear equation.

Q.11x a dy/dx + y x a +() - =+() 2 4 (OCR unclear).v

Solution

The expression as given cannot be parsed; send the ODE in standard notation.

Answer:

Cannot solve — unclear; please re-enter the equation.

Q.12dy/dx = x + ... (OCR: 'dy dx x x x x y =+−+ sin 2').v

Solution

Please supply the exact formula for dy/dx (including placement of parentheses); I'll then solve.

Answer:

Cannot solve — please provide clear form.

Q.13x dy/dx + y x x += log (OCR ambiguous).v

Solution

Provide the full differential equation (for example: x dy/dx + y = x ln x) and I will give the solution.

Answer:

Cannot solve — equation ambiguous in OCR. Please retype clearly.

Q.15dy/dx = (y + x + 3)/ (1 + 2?) OCR: 'dy dx y x x += 3 1 2, given that y = 2 when x = 1'.v

Solution

The OCR is ambiguous about coefficients and denominators. Send the precise equation and initial condition.

Answer:

Cannot solve — equation text unclear. Please provide exact form of the ODE.

EXERCISE 10.8EXERCISE 10.810 questions

Q.1The rate of increase in the number of bacteria is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours (starting from N_0).v

Solution

Model N' = kN => N(t)=N_0 e^{kt}. Given N(5)=3N_0 => e^{5k}=3 => k=(1/5)\ln 3. Then N(10)=N_0 e^{10k}=N_0 e^{2\ln3}=N_0 3^2 =9N_0.

Answer:

N(10)=N_0 \cdot 9

Q.2Find population P(t) given growth proportional to population and P(0)=300000, P(40)=400000.v

Solution

P' = kP => P(t)=P_0 e^{kt}. Given P(40)=400000=300000 e^{40k} => e^{40k}=4/3 => k=\tfrac{1}{40}\ln(4/3). Thus P(t)=300000 e^{(\ln(4/3)/40) t} =300000 (4/3)^{t/40}.

Answer:

P(t)=300000 \cdot \bigl(\tfrac{4}{3}\bigr)^{t/40} =300000\,e^{(\ln(4/3)/40) t}

Q.3For circuit with E = R i + L (di/dt), find the current i(t) when E = 0.v

Solution

With E=0: L di/dt + R i = 0 => di/dt = -(R/L) i. Solve: i(t)=C e^{-\tfrac{R}{L}t}. Using initial current i(0)=i_0 gives i(t)=i_0 e^{-R t / L}.

Answer:

i(t)=i(0) e^{-\tfrac{R}{L}t}

Q.4The engine of a motor boat moving at 10 m s^{-1} is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equals the velocity at that time. Find the velocity after 2 seconds of switching off the engine.v

Solution

dv/dt = −v, v(0)=10. Solve: v = 10 e^{-t}. At t=2: v(2)=10 e^{-2} ≈1.353 m/s.

Answer:

v(2)=10e^{-2} m/s ≈ 1.353 m/s

Q.5Suppose a person deposits ₹10,000 in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later?v

Solution

Continuous compounding: A = P e^{rt} with P=10000, r=0.05, t=1.5. So A=10000 e^{0.075} ≈10000×1.07788≈₹10,778.8.

Answer:

A = 10000 e^{0.05×1.5} = 10000 e^{0.075} ≈ ₹10,778.8

Q.6Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present. In a certain sample 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. What percentage of the original radioactive nuclei will remain after 1000 years?v

Solution

N' = −kN ⇒ N(t)=N_0 e^{-kt}. Given N(100)=0.9N_0 ⇒ e^{-100k}=0.9. Then N(1000)=N_0 e^{-1000k}=(e^{-100k})^{10}=0.9^{10}≈0.348678 ⇒ about 34.87%.

Answer:

Remaining fraction after 1000 yr = 0.9^{10} ≈ 0.348678 => 34.8678% remain

Q.7Water at temperature 100°C cools in 10 minutes to 80°C in a room temperature of 25°C. Find (i) the temperature of the water after 20 minutes; (ii) the time when the temperature is 40°C.v

Newton's law: T(t)=25+75 e^{-kt}. From T(10)=80 ⇒ 25+75 e^{-10k}=80 ⇒ e^{-10k}=11/15, so k=-(1/10)ln(11/15).

(i) T(20)=25+75 e^{-20k}=25+75(e^{-10k})^2=25+75(11/15)^2=25+75·121/225=25+40.333…=65.33°C (approx).

(ii) For T=40: 40=25+75 e^{-kt} ⇒ e^{-kt}=1/5 ⇒ t=-(1/k)ln(0.2). Using k=-(1/10)ln(11/15) gives t≈51.9 minutes.

Q.8At 10:00 A.M. a woman took a cup of hot instant coffee from her microwave oven; the temperature then was 180°F; 10 minutes later it was 160°F. Room temperature is 70°F. (i) What was the temperature of the coffee at 10:15 A.M.? (ii) The woman likes to drink coffee when its temperature is between 130°F and 140°F. Between what times should she drink the coffee?v

Model: T(t)=70+110 e^{-kt}. From T(10)=160 ⇒ 70+110 e^{-10k}=160 ⇒ e^{-10k}=9/11, so k=(1/10)ln(11/9).

(i) T(15)=70+110 e^{-15k}=70+110(e^{-10k})^{1.5}=70+110(9/11)^{1.5}≈151.39°F.

(ii) Solve 70+110 e^{-kt}=140 ⇒ e^{-kt}=70/110=0.63636 ⇒ t≈22.5 min. For 130°F: e^{-kt}=60/110=0.54545 ⇒ t≈30.2 min. Thus she should drink between about 10:22.5 A.M. and 10:30.2 A.M. (≈10:22:30 to ≈10:30:12).

Q.9A pot of boiling water at 100°C is removed from a stove at t = 0 and left to cool. After 5 minutes the temperature is 80°C and after another 5 minutes (t = 10) it is 65°C. Determine the temperature of the kitchen.v

Let room temperature be r and T(t)=r+(100−r)e^{-kt}. Put u=e^{-5k}. From T(5)=80 ⇒ r+(100−r)u=80 and T(10)=65 ⇒ r+(100−r)u^2=65.

Subtracting gives (100−r)u(1−u)=15, and from the first equation (100−r)(1−u)=20. Dividing gives u=15/20=0.75. Then (100−r)(1−0.75)=20 ⇒ (100−r)·0.25=20 ⇒ 100−r=80 ⇒ r=20°C.

Q.10A tank initially contains 50 litres of pure water. Starting at t=0 a brine containing 2 grams of dissolved salt per litre flows into the tank at 3 L/min. The mixture flows out at the same rate. Find the amount of salt present in the tank at any time t>0.v

Solution

dS/dt = in − out = 3×2 − 3(S/50) = 6 − (3/50)S. Solve linear: dS/dt + (3/50)S = 6. Integrating factor e^{3t/50}. Solution S=100 + C e^{-3t/50}. With S(0)=0 ⇒ C=−100. Thus S(t)=100(1−e^{-3t/50}).

Answer:

S(t)=100(1−e^{-3t/50}) grams

Choose the correctChoose the correct25 questions

Q.1The order and degree of the differential equation (interpreted as containing derivatives up to order 4, all to first power) are respectively?v

Solution

If the highest derivative present is of order 4 and derivatives appear to first power, the order is 4 and the degree (power of highest derivative when equation is a polynomial in derivatives) is 1.

Answer:

Order = 4, Degree = 1

Q.2The differential equation representing the family of curves y = A cos x + B, where A and B are arbitrary constants, is:v

y = A cos x + B ⇒ y' = −A sin x, y'' = −A cos x. From y'' + y = B (constant). Differentiating to eliminate B gives y''' + y' = 0, which is satisfied by the family.

Hence the ODE with no arbitrary constants is y''' + y' = 0.

Q.3The order and degree of the differential equation sin x + cos x (dx/dy) = − x (dx/dy) is: (1) order 1, degree 2 (2) order 2, degree 2 (3) order 1, degree 1 (4) order 2, degree 1v

Solution

The displayed form suggests a first-order equation linear in dy/dx (no fractional powers). Hence order =1 and degree =1.

Answer:

Option (3): order 1, degree 1

Q.4The order of the differential equation of all circles with centre at (h,k) and radius a is:v

A general circle (x−h)^2+(y−k)^2=a^2 contains three arbitrary constants h, k and a. The differential equation representing this family therefore has order 3.

Q.5The differential equation of the family of curves y = A e^x + B e^{-x} (A,B constants) is:v

Differentiate twice: y'' = A e^x + B e^{-x} = y. Hence y'' − y = 0, which contains no arbitrary constants.

Q.6The general solution of the differential equation dy/dx = y/x is: (1) xy = k (2) y = k log x (3) y = kx (4) log y = kx.v

Solution. Separating variables, dy/y = dx/x. Hence log y = log x + C, so y = kx.

Answer: Option (3): y = kx.

Q.7The solution of the differential equation (OCR unclear: presented as '2 3 x dy dx y − =') represents which family of curves?v

Solution

The OCR is ambiguous. If the equation were dy/dx = y/x (or linear in y/x) the solution would be power-law; if it were linear yielding y = mx + c the solution would be straight lines. Due to ambiguity this answer is tentative.

Answer:

Option (1): straight lines (chosen with low confidence)

Q.8The solution of dy/dx + p(x)y = 0 is: (1) y = ce^{\int p(x)dx} (2) y = ce^{-\int p(x)dx} (3) x = ce^{-\int p(y)dy} (4) x = ce^{\int p(y)dy}.v

Solution

Standard linear homogeneous equation: integrating factor e^{∫ p dx} gives y e^{∫ p dx} = C ⇒ y = C e^{-∫ p dx}.

Answer:

Option (2): y = C e^{-∫ p(x) dx}

Q.9The integrating factor of the differential equation dy/dx + y = 1 + y/x ??? (OCR unclear), use P(x)=1v

Solution

For dy/dx + P(x) y = Q(x), integrating factor μ= e^{∫P(x) dx}. If P(x)=1 then μ=e^{x}.

Answer:

Option (4): e^{x}

Q.10If the integrating factor of dy/dx + P(x) y = Q(x) is μ(x)=x, then P(x)= ?v

Solution

μ(x)=e^{∫P dx}=x ⇒ ∫P dx = ln x ⇒ P(x)=1/x.

Answer:

Option (3): 1/x

Q.11The degree of the differential equation y = x(1 + dy/dx + (1/2)(dy/dx)^2 + (1/(2.3))(dy/dx)^3 + ...) is: (1) 2 (2) 3 (3) 1 (4) 4.v

Solution

The OCR is not clear; degree is the highest power (positive integer) of the highest order derivative when equation is polynomial in derivatives. I have guessed 3 given the displayed terms; this is tentative.

Answer:

Option (2): 3 (chosen with low confidence)

Q.12If p and q are the order and degree of the differential equation (OCR corrupted), then which relation holds between p and q?v

Solution

The OCR text is unclear; sometimes degree may not exist if the equation is not a polynomial in derivatives. I indicate the case 'degree does not exist' as a likely intended choice but this is tentative.

Answer:

Option (4): p exists and q does not exist (chosen with low confidence)

Q.13The solution of the differential equation (OCR unclear) is which of the given forms?v

Solution

The OCR rendering of the equation and options is unclear; cannot reliably determine the intended differential equation or its solution without the correct statement. Please provide a clearer image or text of the question so I can give the exact option and short justification.

Answer:

Answer provided is ambiguous due to OCR; no confident selection.

Q.14 Solve the differential equation dy/dx = 2 x y.

Answer: (1)

Separate variables: dy/y = 2x dx. Integrate: ln|y| = x^2 + C => y = C e^{x^2}.

Q.15 Solve log(dy/dx) = x + y.

Answer: (1)

From log(dy/dx)=x+y => dy/dx = e^{x+y}=e^x e^y. Separate: e^{-y} dy = e^x dx. Integrate: -e^{-y}=e^x + C => e^x + e^{-y}=C.

Q.16 Solve dy/dx = (y/x) - 2.

Answer: (4)

Write dy/dx - (1/x)y = -2. IF μ=e^{∫(-1/x)dx}=1/x. Then (y/x)' = -2/x. Integrate: y/x = -2 ln|x| + C => y = x(C - 2 ln|x|).

Q.17 If differential equation is of the form dy/dx = φ(y/x) (homogeneous), its general solution can be written in terms of φ. (Choose the correct form.)

Answer: (1)

For a homogeneous DE dy/dx = φ(y/x), put v = y/x => y = vx, dy/dx = v + x dv/dx. Then v + x dv/dx = φ(v) => x dv/dx = φ(v)-v. Separate and integrate to obtain an implicit relation which can be expressed as x = k·F(v) i.e. x = k φ(y/x) (form (1)).

Q.18 If sin x is an integrating factor of dy/dx + P(x) y = Q(x), then P(x) = ?

Answer: (4)

If μ = sin x is an integrating factor then μ = e^{∫P dx} => ∫P dx = ln(sin x) => P = d/dx(ln sin x) = cos x / sin x = cot x.

Q.19 The number of arbitrary constants in the general solutions of order n and n+1 are respectively:

Answer: (2)

A differential equation of order n has a general solution containing n arbitrary constants; order n+1 has n+1 arbitrary constants.

Q.20 The number of arbitrary constants in the particular solution of a differential equation of third order is:

Answer: (4)

A particular (specific) solution has no arbitrary constants, so for any order the particular solution contains 0 arbitrary constants.

Q.21 Find an integrating factor for dy/dx + (1/(x+1)) y = f(x).

Answer: (2)

IF μ = e^{∫P dx} with P=1/(x+1) gives μ = e^{ln|x+1|}=|x+1|. Take μ=x+1.

Q.22 If population P has rate of increase proportional to P, then P(t) = ?

Answer: (1)

dP/dt = kP => dP/P = k dt => ln P = kt + C => P = C e^{kt}.

Q.23 If amount P decays at rate proportional to amount remaining, then P(t) = ?

Answer: (2)

dP/dt = -kP => dP/P = -k dt => P = C e^{-k t}.

Q.24 If the solution of the differential equation dy/dx + a x + y + 2 = 0 represents a circle, find a.

Answer: (2)

For the family to represent a circle the quadratic terms in x and y coming from integration must balance appropriately; the standard result gives a = -2. (Answer supplied with low confidence due to ambiguous statement.)

Q.25 The slope dy/dx of a curve y=f(x) is given by dy/dx = 3x^2 and the curve passes through (-1,1). Find the equation of the curve.

Answer: (1)

Integrate: y = ∫3x^2 dx = x^3 + C. Use point (-1,1): 1 = (-1)^3 + C => C=2. So y = x^3 + 2.

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