Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.5 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5

Question 1.

If F is the constant force generated by the motor of an automobile of mass M, its velocity V is given by M \(\frac { dV }{ dt }\) = F – kV, where k is a constant. Express V in terms of t given that V = 0 when t = 0.

Solution:

Given differential equation is M \(\frac { dV }{ dt }\) = F – kV

The given equation can be written as

\(\frac { dV }{ F-kV }\) = \(\frac { dt }{ M }\)

Now Integrating, we get

Initial condition:

Given V = 0 When t = 0

C = e^{\(\frac { k(0) }{ M }\)} [F – k (0)]

= e° [F – 0]

C = F

∴ (1) ⇒ F = (F – kV)e^{\(\frac { kt }{ M }\)}

Question 2.

The velocity v, of a parachute falling vertically satisfies the equation v\(\frac { dv }{ dx }\) = g(1 – \(\frac { v^2 }{ k^2 }\)) where g and k are constants. If v and are both initially zero, find v in terms of x.

Solution:

Initial condition:

Given V = 0 when x = 0, we get

k²(0)² = Ce^{\(\frac { -2g(0) }{ k^2 }\)}

k² = Ce°

k² = C

(1) ⇒ k² – v² = k² e^{\(\frac { -2gx }{ k^2 }\)}

k² – k² e^{\(\frac { -2gx }{ k^2 }\)} = v²

k² [1 – e^{\(\frac { -2gx }{ k^2 }\)}] = v²

Question 3.

Find the equation of the curve whose slope is \(\frac { y-1 }{ x^2+x }\) and which passes through the point (1, 0).

Solution:

1 = A(x + 1) + B(x)

Put x = -1; Put x = 0

1 = A (0) + B (-1) ; 1 = A(0 + 1) + B(0)

1 = -B; 1 = A

B = -1; A = 1

∴ \(\frac { 1 }{ x^2+x }\) = \(\frac { 1 }{ x }\) + \(\frac { 1 }{ x+1 }\) ……. (2)

Substituting equation (2) in equation (1), we get

\(\frac { dy }{ y-1 }\) = \(\frac { dx}{ x }\) + \(\frac { dx }{ x+1 }\)

Taking integrating on both sides, we get

log (y – 1) = log x – log (x + 1) + log C

log (y – 1) = log C + log x – log (x + 1)

= log Cx – log (x + 1)

Question 4.

Solve the following differential equations:

(i) \(\frac { dy }{ dx }\) = \(\sqrt { 1-y^2 }{ 1-x^2 }\)

(ii) ydx + (1 + x²) tan^{-1} x dy = 0

(iii) sin \(\frac { dy }{ dx }\) = a, y (0) = 1

(iv) \(\frac { dy }{ dx }\)e^{x+y} + x³, e^{y}

(v) (e^{y} + 1) cos x dx + e^{y} sin x dy = 0

(vi) (ydx – xdy) cot (\(\frac { x }{ y }\)) = ny² dx

(vii) \(\frac { dy }{ dx }\) – x\(\sqrt { 25-x^2 }\) = 0

(viii) x cos y dy = e^{x} (x log x + 1) dx

(xi) tan y \(\frac { dy }{ dx }\) = cos (x + y) + cos (x – y)

(x) \(\frac { dy }{ dx }\) = tan² (x + y)

Solution:

(i) \(\frac { dy }{ dx }\) = \(\sqrt { 1-y^2 }{ 1-x^2 }\)

The equation can be written as

\(\frac { dy }{ \sqrt {1-y^2} }\) = \(\frac { dx }{ \sqrt {1-x^2} }\)

Taking Integration on both sides, we get

∫ \(\frac { dy }{ \sqrt {1-y^2} }\) = ∫ \(\frac { dx }{ \sqrt {1-x^2} }\)

sin^{-1}y = sin^{-1} x + C

(ii) ydx + (1 + x²) tan^{-1} x dy = 0

ydx = – (1 + x²) tan^{-1} x dy

Take t = tan^{-1} x

dt = \(\frac { 1 }{ 1+x^2 }\) dx

The equation can be written as

\(\frac { dx }{ (1+x^2)tan^{-1}x }\) = –\(\frac { dy }{ y }\)

\(\frac { dt }{ t }\) = –\(\frac { dy }{ y }\)

Taking Integration on both sides, we get

∫ \(\frac { dt }{ t }\) = ∫ \(\frac { dy }{ y }\)

log t = – log y + log C

log (tan^{-1} x) = – log y + log C

log (y(tan^{-1} x)) + log y = log C

y tan^{-1} x = C

(iii) sin \(\frac { dy }{ dx }\) = a, y (0) = 1

sin \(\frac { dy }{ dx }\) = a

sin \(\frac { dy }{ dx }\) = sin^{-1} (a)

The equation can be written as

dy = sin^{-1} (a) dx

Taking integration on both sides, we get

∫ dy = ∫ sin^{-1} (a) dx

y = sin^{-1} a ∫ dx

y = sin^{-1} (a) x + C ……… (1)

Initial condition:

Since y (0) = 1, we get

y = sin^{-1} (a) x + C

1 = sin^{-1} (a) (0) + C

0 + C = 1

C = 1

Equation (1) ⇒ y = sin^{-1} (a) x + 1

y – 1 = sin^{-1} (a) x

\(\frac { y-1 }{ x }\) = sin^{-1} (a)

sin(\(\frac { y-1 }{ x }\)) = a

(iv) \(\frac { dy }{ dx }\)e^{x+y} + x³, e^{y}

\(\frac { dy }{ dx }\)e^{x+y} + x³, (e^{y})

= e^{y} [e^{x} + x³]

\(\frac { dy }{ e^y }\) = dx(e^{x} + x³)

The equation can be written as

\(\frac { dy }{ e^y }\) = (e^{x} + x³) dx

Taking integration on both sides, we get

∫ e^{-y} dy = ∫ (e^{x} + x³) dx

\(\frac { e^y }{ -1 }\) = e^{x} + \(\frac { x^4 }{ 4 }\) + C

[Where – C = C, Which is also constant].

∴ e^{x} + e^{-y} + \(\frac { x^4 }{4 }\) = -C = C

∴ e^{x} + e^{-y} + \(\frac { x^4 }{4 }\) = C

(v) (e^{y} + 1) cos x dx + e^{y} sin x dy = 0

Solution:

(e^{y} + 1) cos x dx + e^{y} sin x dy = 0

e^{y} sin x dy = – (e^{y} + 1) cos x dx

log (e^{y} + 1) = – log sin x + log c

log [(e^{y} + 1) + log sin x = log c

log (e^{y} +1) sin x] = log c

(e^{y}+ 1) sin x = c

(vi) (ydx – xdy) cot (\(\frac { x }{ y }\)) = ny² dx

equation can be written as

Substituting these values in equation (1), we get

dt cot t = ndx

cot t dt = ndx

Taking integration on both sides, we get

∫ cot t dt = n∫ dx

log (sin t) = n x + c

sin t = e^{nx+c}

∴ sin(\(\frac { x }{ y }\)) = e^{nx+c} [∵ t = \(\frac { x }{ y }\) ]

(vii) \(\frac { dy }{ dx }\) – x\(\sqrt { 25-x^2 }\) = 0

The equation can be written as

\(\frac { dy }{ dx }\) – x\(\sqrt { 25-x^2 }\) …….. (1)

Take 25 – x² = t

-2x dx = dt

x dx = –\(\frac { dt }{ 2 }\)

Substituting these values in equation (1), we get

dy = x\(\sqrt { 25-x^2 }\) dx

dy = -√t \(\frac { dt }{ 2 }\)

Taking integration on both sides, we get

∫ dy = –\(\frac { dt }{ 2 }\) ∫ t^{\(\frac { 1 }{ 2 }\)} dt

(viii) x cos y dy = e^{x} (x log x + 1) dx

The equation can be written as

Substituting in (1), we get

sin y = e^{y} log x + C

(ix) tan y \(\frac { dy }{ dx }\) = cos (x + y) + cos (x – y)

The equation can be written as

tan y \(\frac { dy }{ dx }\) = cos (x + y) + cos (x – y)

[W.K.T cos (A + B) + cos (A – B) = 2 cos A cos B

Here A = x, B = y]

∴ tan y\(\frac { dy }{ dx }\) = 2 cos x cos y

\(\frac { tany }{ cosy }\) dy = 2 cos x dx

Taking integration on both sides, we get

∫ \(\frac { tany }{ cosy }\) dy = 2 ∫ cos x dx

2 ∫ tan y sec y dy = 2 ∫ cos x dx

sec y = 2 sin x + C

(x) \(\frac { dy }{ dx }\) = tan² (x + y)