## MP Board Class 12th Maths Important Questions Chapter 9 Differential Equations

### Differential Equations Important Questions

**Differential Equations Objective Type Questions**

Question 1.

Choose the correct answer:

Question 1.

Degree of the differential equation

\(\frac { d^{ 2 }y }{ d^{ 2 }x } \) + x^{2} \(\frac{dy}{dx}\) = e^{x} is:

(a) 1

(b) 2

(c) 3

(d) does not exist

Answer:

(c) 3

Question 2.

Solution of differential equation (1 + x) y dx + (1 – y) xdy = 0 is:

(a) log xy + x + y = c

(b) logxy + x – y = c

(c) log xy – x – y = c

(d) log xy – x + y = c.

Answer:

(b) logxy + x – y = c

Question 3.

The differential equation of all circles which passes through the origin whose center lie on the A – axis is:

(a) x^{2} = y^{2}+ xy \(\frac{dy}{dx}\)

(b) x^{2} = y^{2} + 3xy \(\frac{dy}{dx}\)

(c) y^{2} = x^{2} + 2xy \(\frac{dy}{dx}\)

(d) y^{2} = x^{2} – 2xy \(\frac{dy}{dx}\)

Answer:

(c) y^{2} = x^{2} + 2xy \(\frac{dy}{dx}\)

Question 4.

Solution of the differential equation \(\frac{dy}{dx}\) + y = e^{x}, y(o) = 0 is:

(a) y = e^{-x}(x – 1)

(b) y = xe^{x}

(c) y = xe^{-x} + 1

(d) y = xe^{-x}

Answer:

(d) y = xe^{-x}

Question 5.

The straight line which satisfies the differential equation \(\frac{dy}{dx}\) = m and cuts an intercept 3 on the positive y – axis is:

(a) y = mx + c

(b) y = mx +3

(c) y = mx – 3

(d) y = – mx + 3.

Answer:

(b) y = mx +3

Question 2.

Fill in the blanks:

- The corresponding differential equation of the equation x
^{2}+ y^{2}= a^{2}is …………………………… - The differential equation of the curve y = e
^{cx}is …………………….. where c is arbitrary constant. - The integration factor of the linear differential equation \(\frac{dy}{dx}\) + Py = Q is …………………………….
- In the linear differential equation \(\frac{dy}{dx}\) + Py = Q. P and Q are …………………………….
- The form of differential equation (x + y + 1) dy = dx is ……………………………
- Solution of the differential equation e
^{-x+y}\(\frac{dy}{dx}\) = 1 is ……………………………

Answer:

- y \(\frac{dy}{dx}\) + x = 0
- x \(\frac{dy}{dx}\) = y log y
- e
^{∫}^{pdx} - constant
- linear differential equation
- e
^{-y}= e^{-x}+ c

Question 3.

Write True/False:

- Order of differential equation y = x \((\frac { dy }{ dx } )^{ 2 }\) + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) is 2.
- Degree of differential equation \((\frac { d^{ 3 }y }{ dx^{ 3 } } )^{ 4/5 }\) – 2 \((\frac { dy }{ dx } )\) \((\frac { d^{ 2 }y }{ dx^{ 2 } } )^{ 2 }\) = 0 is 5.
- The integration factor of linear differential equation x \(\frac{dy}{dx}\) – y = 2x
^{2}is e^{-x} - Solution of differential equation dy = sin xdx is y + cos x – c = 0.
- Solution of differential equation ydx + (x – y
^{3}) dy = 0 is xy = \(\frac { y^{ 4 } }{ 4 } \) + c.

Answer:

- True
- False
- False
- True
- True

Question 4.

Write the answer is one word/sentence:

- Write the integration factor of differential equation (1 + y
^{2}) + (2xy – cot y) \(\frac{dy}{dx}\) = 0. - Find the solution of the differential equation (1 + x
^{2}) dy = (1 + y^{2}) dx. - Find the sum of order and degree of the differential equation y = x \((\frac { dy }{ dx } )^{ 3 }\) + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \)
- Find the integration factor of the differential equation x logx \(\frac{dy}{dx}\) + y = 2 log x
- Find he solution of the differential equation \(\frac{dy}{dx}\) + \(\frac{1}{x}\) = \(\frac { e^{ y } }{ x^{ 2 } } \)

Answer:

- 1 + y
^{2} - x – y = c(1 + xy)
- 3
- log x
- 2xe
^{-y}= cx^{2}+ 1.

**Differential Equations Very Short Answer Type Questions**

Question 1.

Find the order and degree of \(\frac{dy}{dx}\) + y = e^{-x}?

Answer:

1, 1

Question 2.

Find the degree and order of \((\frac { dy }{ dx } )^{ 3 }\) = \(\sqrt { 1+(\frac { dy }{ dx } )^{ 2 } } \)?

Answer:

1, 6

Question 3.

Find the degree and order of \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + \(\sqrt { 1+(\frac { dy }{ dx } )^{ 3 } } \) = 0?

Answer:

2, 2

Question 4.

Find the differential equation corresponding to equation of circle x^{2} + y^{2} = a^{2}?

Answer:

y \(\frac{dy}{dx}\) + x = 0

Question 5.

Find the differential equation of line y = mx + c?

Answer:

\(\frac{dy}{dx}\) = m

Question 6.

Solve the differential equation \(\frac{dy}{dx}\) = 4y?

Answer:

y = c.e^{4x}

Question 7.

Solve the differential equation x^{2} \(\frac{dy}{dx}\) = 2?

Answer:

y = c – \(\frac{2}{x}\)

Question 8.

Find the solution of differential equation dy = sinxdx?

Answer:

y + cos x = c

Question 9.

Find the solution of differential equation \(\frac{dy}{dx}\) + Px = Q?

Answer:

xe∫^{pdy}. dy + c

Question 10.

Solve the differential equation (1 – y^{2}) \(\frac{dy}{dx}\) + yx = ay?

Answer:

\(\frac { 1 }{ \sqrt { 1-y^{ 2 } } } \)

**Differential Equations Short Answer Type Questions**

Question 1.

Solve the differential equation xlog xdy – ydx = 0?

Solution:

x log x dy – ydx = 0 (given)

⇒ x log x = ydx

⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = \(\frac{1}{xlogx}\) dx

⇒ ∫\(\frac{1}{y}\) dy = ∫\(\frac{1}{xlogx}\) dx

⇒ log y = ∫\(\frac{1}{t}\) dt , (let log x = t ⇒ \(\frac{1}{x}\) dx = dt)

⇒ log y = log t + log c

⇒ log y = log log x + log c

Question 2.

Solve the differential equation \(\frac{dy}{dx}\) = e^{x-y} + x.e^{-y}?

Solution:

\(\frac{dy}{dx}\) = e^{x-y} + x.e^{-y} (given)

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

⇒ e^{y}dy = (e^{x} + x) dx

Integrating both sides,

∫e^{y}dy = ∫(e^{x} + x) dx

e^{y} = e^{x} + \(\frac { x^{ 2 } }{ 2 } \) + c

Question 3.

Prove that solution of y = 4sin 3x is \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + 9y = 0?

Solution:

y = 4 sin 3x (given)

Differentiating with respect to x,

∴\(\frac{dy}{dx}\) = 12 cos 3x

Again differentiating with respect to x,

\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = – 36 sin 3x = – 9 × 4 sin 3x

⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = -9y, [from eqn.(1)]

⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + 9y = 0. Proved.

Question 4.

Solve the differential equation \(\frac{dy}{dx}\) = sec^{2} x + 2x?

Solution:

\(\frac{dy}{dx}\) = sec x (sec x + tan x) (given)

⇒ dy = (sec^{2} x + sec x tan x ) dx

⇒ ∫dy = ∫sec^{2} dx + ∫sec x tan x dx

∴y = tan x + sec x + c.

Question 5.

Solve the differential equation \(\frac{dy}{dx}\) = sec^{2} x + 3x^{2}?

Solution:

\(\frac{dy}{dx}\) = sec^{2} x + 3x^{2} (given)

⇒ dy = (sec^{2} x + 3x^{2}) dx

⇒ ∫dy = ∫sec^{2} xdx + 3∫x^{2} dx

⇒ y = tan x + \(\frac { 3x^{ 3 } }{ 3 } \) + c

⇒ y = tan x + x^{3} + c.

Question 6.

Solve the differential equation \(\frac{dy}{dx}\) = sec^{2} x + 2x?

Solution:

Solve as Q.No . 5

Question 7.

Solve the differential equation \(\frac{dy}{dx}\) = (3x^{2} x + 2)?

Solution:

\(\frac{dy}{dx}\) = (3x^{2} + 2) (given)

⇒ dy = (3x^{2} + 2) dx

⇒ ∫dy = ∫(3x^{2} + 2) dx

⇒ y = 3 × \(\frac { x^{ 3 } }{ 3 } \) + 2x = c = x^{2} + 2x + c.

Question 8.

Solve the differential equation x^{2} \(\frac{dy}{dx}\) = 2?

Solution:

x^{2} \(\frac{dy}{dx}\) = 2 (given)

⇒ dy = 2.x^{-2} dx

⇒ ∫dy = 2∫x^{-2} dx

⇒ y = 2 \((\frac { -1 }{ x } )\) + c

Question 9.

Solve the differential equation \(\frac{dy}{dx}\) = x^{3} + sin 4x?

Solution:

\(\frac{dy}{dx}\) = x^{3} + sin 4x (given)

⇒ dy = (x^{3} + sin 4x) dx

⇒ ∫dy = ∫x^{3} dx + ∫sin 4x dx

⇒ y = \(\frac { x^{ 4 } }{ 4 } \) + \((\frac { -cos4x }{ 4 } )\) + c

⇒ y = \(\frac { x^{ 4 } }{ 4 } \) – \(\frac { -cos4x }{ 4 } \) + c

Question 10.

Solve the differential equation \(\frac{dy}{dx}\) + 2x = e^{3x}?

Solution:

\(\frac{dy}{dx}\) + 2x = e^{3x} (given)

⇒ \(\frac{dy}{dx}\) = e^{3x} – 2x

⇒ dy = (e^{3x} – 2x) dx

⇒ ∫dy = ∫^{3x} dx – 2∫x dx

⇒ y = e^{3x}. \(\frac{1}{3}\) e^{3x} – x^{2} + c

Question 11.

Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \) (given)

⇒ \(\frac { 1 }{ cos^{ 2 }y } \) dy = \(\frac { 1 }{ sin^{ 2 }x } \) dx

⇒ sec^{2} ydy = cosec^{2} xdx

⇒ ∫sec^{2} ydy = ∫cosec^{2} xdx

⇒ tan y = – cot x + c

Question 12.

Solve the differential equation (x^{2} + 1) \(\frac{dy}{dx}\) = 1?

Solution:

Question 13.

Solve the differential equation \(\frac{dy}{dx}\) = sin x sin y?

Solution:

\(\frac{dy}{dx}\) = sin x sin y

⇒ cosec y dy = sin x dx

On integrating,

– log_{e} (cosec y + cot y) = – cos x + c

⇒ cos x – log_{e} (cosec y + cot y) = c

Question 14.

Solve the differential equation \(\frac{dy}{dx}\) = y sin x?

Solution:

\(\frac{dy}{dx}\) = y sin x

⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = sin x

⇒ ∫\(\frac{1}{y}\) dy = ∫sin x dx

⇒ log y = – cos x + c

Question 15.

Solve the differential equation \(\frac{dy}{dx}\) = x cos x?

Solution:

Given:

\(\frac{dy}{dx}\) = x cos x

⇒ dy = x cos x dx

⇒ ∫dy = ∫x cos x dx

⇒ y = xsin x – ∫1. sin x dx + c

⇒ y = xsin x + cos x + c

Question 16.

Solve the differential equation \(\frac{dy}{dx}\) = x^{3} + sin 4x?

Solution:

\(\frac{dy}{dx}\) = x^{3} + sin 4x (given)

⇒ dy = (x^{3} + sin 4x) dx

⇒ ∫dy = ∫x^{3} dx + ∫sin 4x dx

⇒ y = \(\frac { x^{ 4 } }{ 4 } \) + \((\frac { -cos4x }{ 4 } )\) + c

⇒ y = \(\frac { x^{ 4 } }{ 4 } \) – \(\frac { cos4x }{ 4 } \) + c

Question 17.

Solve the differential equation \(\frac{dy}{dx}\) + 2x = e^{3x}?

Solution:

Question 18.

Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \)?

Solution:

\(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \)

⇒ \(\frac { 1 }{ cos^{ 2 }y } \) dy = \(\frac { 1 }{ sin^{ 2 }x } \) dx

⇒ sec^{2} ydy = cosec^{2} xdx

∴∫sec^{2} ydy = ∫cosec^{2} xdx + c

tan y = – cotx + c

Question 19.

Solve the differential equation \(\frac{dy}{dx}\) = 1 – x + y – xy?

Solution:

\(\frac{dy}{dx}\) = 1 – x + y – xy (given)

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

⇒ \(\frac{dy}{dx}\) = (1 – x) (1 + y)

⇒ \(\frac{dy}{1+y}\) = (1 – x) dx

⇒ ∫\(\frac{dy}{1+y}\) = ∫(1 – x) dx

⇒ log_{e} (1 + y) = x – \(\frac { x^{ 2 } }{ 2 } \) + c

Question 20.

Solve the differential equation \(\frac{dy}{dx}\) = (1 + x) (1 + y^{2})?

Solution:

\(\frac{dy}{dx}\) = (1 + x) (1 + y^{2}) (given)

⇒ \(\frac { 1 }{ 1+y^{ 2 } } \) dy = (1 + x) dx

Integrating both sides,

⇒ tan^{-1}y = x + \(\frac { x^{ 2 } }{ 2 } \) + c.

Question 21.

Solve the differential equation:

\(\frac{dy}{dx}\) = cot^{2} x?

Solution:

\(\frac{dy}{dx}\) = cot^{2} x (given)

⇒ dy = cot^{2} x

⇒ ∫dy = ∫cot^{2} x dx

⇒ y = ∫(cosec^{2} x – 1) dx

⇒ y = – cot x – x + c

**Differential Equations Long Answer Type Questions – II**

Question 1.

(A) Solve the differential equation \(\frac{dy}{dx}\) + y tan x = sec x?

Solution:

\(\frac{dy}{dx}\) + y tan x = sec x (given)

Comparing the equation with \(\frac{dy}{dx}\) + Py = Q,

P = tan x, Q = sec x

∴L.F. = e^{∫P dx} = e^{∫}^{tanx dx} = e^{log secx}

⇒ L.F. = sec x

Applying formula

y × (1.F.) = ∫Q × (I.F.) dx + c

⇒ ∫sec^{2} x dx + c

⇒ y sec x = tan x + c

(B) Solve the differential equation \(\frac{dy}{dx}\) + y tan x = sin x?

Solution:

Solve as Q.No.1 (A).

Question 2.

Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { \sqrt { 1-y^{ 2 } } }{ \sqrt { 1-x^{ 2 } } } \)?

Solution:

Question 3.

Solve the differential equation 3x^{2} dy = (3xy + y^{2})dx?

Solution:

3x^{2} dy = (3xy + y^{2}) dx (given)

Putting in eqn. (1),

⇒ – \(\frac{1}{v}\) = \(\frac{1}{3}\) log x + c

∴- \(\frac{1}{y}\) = \(\frac{1}{3}\) log x + c

Question 4.

Solve the differential equation (1 + x^{2}) \(\frac{dy}{dx}\) + 2xy = 4x^{2}?

Solution:

The required solution will be:

Question 5.

Solve the differential equation (1 + x^{2}) \(\frac{dy}{dx}\) + 2xy = cos x?

Solution:

(1 + x^{2}) \(\frac{dy}{dx}\) + 2xy = cos x (given)

⇒ \(\frac{dy}{dx}\) + \(\frac { 2x }{ (1+x^{ 2 }) } \)

Comparing with \(\frac{dy}{dx}\) + Py = Q,

The required solution of eqn. (1) is:

Question 6.

Marginal cost price of making anything is given by equation c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x. Find the total cost price c(x) for making it? [Given c (0) = 100]

Solution:

c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x (given)

Integrating both sides

∫c'(x) dx = ∫ (2 + 0.15 x) dx

c(x) = 2x + 0.15 \(\frac { x^{ 2 } }{ 2 } \) + A …………. (1)

If x = 0

c(0) = 2 × 0 + \(\frac { 0.15 }{ 2 } \) × 0^{2} + A

⇒ c(0) = A

∴ A = 100, [∵ c(0) = 100]

Putting in eqn. (1),

c(x) = 2x + 0.075 x^{2} + 100

Question 7.

Solve the differential equation x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0?

Solution:

x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0 (given)

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

Integrating both sides,

Question 8.

Solve the differential equation (x + y + 1) \(\frac{dy}{dx}\) = 1?

Solution:

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

⇒ \(\frac{dy}{dx}\) = x + y + 1

⇒ \(\frac{dy}{dx}\) – x = y + 1

Comparing the equation with \(\frac{dx}{dy}\) + Px = Q?

P = – 1 and Q = y + 1

∴I.F. = e^{∫}^{pdy} = e^{-∫}^{dy} = e^{-y}

Here the required solution is:

⇒ x.e^{∫}^{pdy} = ∫e^{∫}^{pdy}. Qdy

⇒ x.e^{-y} = ∫e^{-y} (y + 1) dy + c

⇒ x.e^{-y} = – (y + 1)e^{-y} – ∫1(-e^{-y}) dy + c

⇒ x = -(y + 1) -1 + ce^{y}

x + y + 2 = ce^{y}

Question 9.

Solve the differential equation sec^{2}x tany dx + sec^{2} y tan xdy = 0?

Solution:

sec^{2}x tany dx + sec^{2} y tan xdy = 0 (given)

⇒ log y = – log x + log c

⇒ log x + log y = log c

⇒ log xy = log c

⇒ xy = c

Question 10.

Solve the differential equation \(\frac{dy}{dx}\) = y tanx – 2 sin x?

Solution:

\(\frac{dy}{dx}\) – y tanx = – 2 sin x (given)

Comparing with \(\frac{dy}{dx}\) + Py = Q,

P = – tan x, Q = – 2 sin x

L.F. = e^{∫}^{pdx} = e^{-∫tanxdx}

= e^{logecosx} = cos x

The required solution is:

y.(I.F.) = ∫Q.I.F. dx + c

⇒ ycos x = -2∫sin x cos x dx + c

⇒ ycosx = -∫sin2xdx + c

⇒ ycosx = \(\frac { cos2x }{ 2 } \) + c.

Question 11.

Solve the differential equation \(\frac{dy}{dx}\) + 2y = 4x?

Solution:

Solution:

\(\frac{dy}{dx}\) + 2y = 4x

Comparing with \(\frac{dy}{dx}\) + Py = Q,

P = 2, Q = 4x

I.F. = e^{∫}^{pdx} = e^{∫2dx} = e^{2x}

The required differential solution is:

y. (I.F) = ∫Q.I.F. dx + c

⇒ ye^{2x} = ∫4xe^{2x} dx + c

Question 12.

Solve the differential equation cos^{2} x \(\frac{dy}{dx}\) + y = 2?

Solution:

cos^{2} x \(\frac{dy}{dx}\) + y = 2

⇒ \(\frac{dy}{dx}\) + sec^{2} x. y = 2 sec^{2} x

Comparing with \(\frac{dy}{dx}\) + P y = Q,

P = sec^{2} x, Q = 2 sec^{2} x

I.F. = e^{∫sec2 xdx }

Required solution is:

y. (I.F) = ∫Q.I.F. dx + c

⇒ y.e^{tanx} = ∫2 sec^{2} x. e^{tanx} dx + c

⇒ y.e^{tanx} = 2.∫e^{t} dt + c, (Let tan x = t ⇒ sec^{2} xdx = dt)

= 2.e^{tanx} + c

⇒ e^{tanx} (y – 2) = c.

Question 13.

Solve the differential equation cos x \(\frac{dy}{dx}\) + y = sin x?

Solution:

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

⇒ \(\frac{dy}{dx}\) + sec x.y = tan x

Comparing with \(\frac{dy}{dx}\) + Py = Q,

P = secx, Q = tanx

L.F. = e^{∫secxdx}

= e^{loge(sec x + tan x)} dx + c

Required Solution y.I.F. = ∫Q.I.F. dx + c

⇒ y. (sec x + tan x) = ∫tan x.(sec x + tan x) dx + c

= ∫sec x tan x dx + ∫tan^{2} xdx + c

= sec x + ∫(sec^{2} x – 1) dx = sec x + tan x = – x + c.

Question 14.

Solve the differential equation (1 + y^{2}) dx = (tan^{-1} y – x) dy? (CBSE 2015)

Solution:

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

The required solution is:

Question 15.

Solve the differential equation (1 + y^{2}) + ( x – e^{tan-1 y}) \(\frac{dy}{dx}\) = 0? (CBSE 2016)

Solution:

The given equation is:

Comparing with \(\frac{dx}{dy}\) + Px = Q,

The required solution is:

Put tan^{-1} y = t,

Question 16.

Solve the differential equation (1 + x^{2}) \(\frac{dy}{dx}\) + 2xy = \(\frac { x }{ 1+x^{ 2 } } \). Where, y = 0 and x = 1? (NCERT)

Solution:

Comparing with \(\frac{dy}{dx}\) + Py = Q,

The required solution is:

y = 0 and x = 1 (given)

0 (1 + 1)^{2} = tan^{-1} 1 + c

⇒ 0 = \(\frac { \pi }{ 4 } \) + c

⇒ c = – \(\frac { \pi }{ 4 } \)

From eqn. (2),

y. (1 + x^{2}) = tan^{-1} x – \(\frac { \pi }{ 4 } \).

Question 17.

Solve the differential equation \(\frac{dy}{dx}\) + y cot x = 4x cosec x. Given: y = 0 and x = \(\frac { \pi }{ 2 } \)? (NCERT; CBSE, 2012)

Solution:

\(\frac{dy}{dx}\) + y cot x = 4x cosec x (given)

Comparing with eqn. (1) \(\frac{dy}{dx}\) + Py = Q,

P = cot x, Q = 4x cosec x

I.F. = e^{∫}^{pdx}

= e^{∫}^{cot xdx} = e^{log(sinx)} = sin x

The required solution is:

y.I.F = ∫I.F. × Qdx

⇒ y.sin x = ∫sin x × 4x cosec x dx

⇒ y sin x = 4∫\(\frac { xsinx }{ sinx } \) dx

⇒ y sin x = 4∫xdx

⇒ ysin x = \(\frac { 4x^{ 2 } }{ 2 } \) + c

⇒ y sin x = 2x^{2} + c

When x = \(\frac { \pi }{ 2 } \) and y = 0,

0 (sin \(\frac { \pi }{ 2 } \)) = 2 ( \(\frac { \pi }{ 2 } \) ) ^{2} + c

⇒ 0 = \(\frac { 2\pi ^{ 2 } }{ 4 } \) + c

∴ c = \(\frac { -\pi ^{ 2 } }{ 2 } \)

Putting in eqn. (1),

y. sin x = 2x^{2} – \(\frac { -\pi ^{ 2 } }{ 2 } \).