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NCERT Book Solutions (Class - 6 to Class - 12)

## Differential-Equations

Exercise 9.6

For each of the differential equations given in Exercises 1 to 12, find the general solution:

Question :1:
dy/dx + 2y = sin x

The given differential equation is:
dy/dx + 2y = sin x
This is in the form of dy/dx + py = Q, where p = 2 and Q = sin x
Now, I.F = eꭍp dx = eꭍ2 dx = e2x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y e2x = ꭍ sin x * e2x dx + C   ………..1
Let I = ꭍ sin x * e2x dx
⇒ I = sin x * ꭍ e2x dx - ꭍ[d(sin x/dx * ꭍ e2x dx]dx
⇒ I = sin x * e2x/2 - ꭍ [cos x * e2x/2]dx
⇒ I = sin x * e2x/2 – (1/2) * cos x * ꭍ e2x dx - ꭍ[d(cos x/dx * ꭍ e2x dx]dx
⇒ I = sin x * e2x/2 – (1/2) * [cos x * e2x/2 - ꭍ {(-sin x) * e2x/2]dx]
⇒ I = (sin x * e2x)/2 – (cos x * e2x)/4 – I/4
⇒ I + I/4 = e2x (2 sin x – cos x)/4
⇒ 5I/4 = e2x (2 sin x – cos x)/4
⇒ I = e2x (2 sin x – cos x)/5
Therefore, equation 1 becomes:
⇒ y e2x = e2x (2 sin x – cos x)/5 + C
⇒ y = (2 sin x – cos x)/5 + C e-2x
This is the required general solution of the given differential equation.

Question :2:
dy/dx + 3y = e-2x

The given differential equation is:
dy/dx + 3y = e-2x
This is in the form of dy/dx + py = Q, where p = 3 and Q = e-2x
Now, I.F = eꭍp dx = eꭍ3 dx = e3x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ ye3x = ꭍ (e-2x * e3x) dx + C
⇒ ye3x = ꭍ ex dx + C
⇒ ye3x = ex + C
⇒ y = e-2x + Ce-3x
This is the required general solution of the given differential equation.

Question :3:
dy/dx + y/x = x2

The given differential equation is:
dy/dx + y/x = x2
This is in the form of dy/dx + py = Q, where p = 1/x and Q = x2
Now, I.F = eꭍp dx = eꭍ dx/x = elog x = x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y * x = ꭍ (x2 * x )dx + C
⇒ xy = ꭍ x3 dx + C
⇒ xy = x4/4 + C
This is the required general solution of the given differential equation.

Question :4:
dy/dx + y * sec x = tan x (0 ≤ x ≤ π/2)

The given differential equation is:
dy/dx + y * sec x = tan x
This is in the form of dy/dx + py = Q, where p = sec x and Q = tan x
Now, I.F = eꭍp dx = eꭍ sec x dx = elog (sec x + tan x) = sec x + tan x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y(sec x + tan x) = ꭍ tan x(sec x + tan x)dx + C
⇒ y(sec x + tan x) = ꭍ tan x * sec x dx + ꭍ tan2 x dx + C
⇒ y(sec x + tan x) = sec x + ꭍ(sec2 x – 1) dx + C
⇒ y(sec x + tan x) = sec x + tan x – x + C
This is the required general solution of the given differential equation.

Question :5:
0π/2 cos 2x dx

Let I = ꭍ0π/2 cos 2x dx
⇒ I = [sin 2x /2 0] π/2
⇒ I = (1/2) * [sin π – sin 0]
⇒ I = 0
Hence, ꭍ0π/2 cos 2x dx = 0

Question :6:
x * dy/dx + 2y = x2 log x

The given differential equation is:
dy/dx + 2y/x = x2 log x
This is in the form of dy/dx + py = Q, where p = 2/x and Q = x log x
Now, I.F = eꭍp dx = eꭍ 2/x dx = e2 log x = x2
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y * x2 = ꭍ (x log x * x2)dx + C
⇒ x2y = ꭍ x3 log x dx + C
⇒ x2y = log x * ꭍ x3 dx - ꭍ[d(log x)/dx * ꭍ x3 dx]dx + C
⇒ x2y = log x * x4/4 - ꭍ[ 1/x *  x4/4]dx + C
⇒ x2y = log x * x4/4 – (1/4)ꭍ x3 + C
⇒ x2y = log x * x4/4 – (1/4) * x4/4 + C
⇒ x2y = x4 (4log x – 1)/16 + C
⇒ y = x2 (4log x – 1)/16 + Cx-2
This is the required general solution of the given differential equation.

Question :7:
x * log x dy/dx + y = 2/x log x

The given differential equation is:
x * log x dy/dx + y = 2/x log x
⇒ dy/dx + y/x log x = 2/x2
This is in the form of dy/dx + py = Q, where p = 1/x log x and Q = 2/x2
Now, I.F = eꭍp dx = eꭍ 1/x log x dx = elog(log x) = log x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y * log x = ꭍ(2/ x2 * log x)dx + C
⇒ y * log x = 2 * ꭍ log x/x2 dx + C
⇒ y * log x = 2[ log x * ꭍ 1/x2 dx  - ꭍ{d(log x)/dx * ꭍ1/x2 dx} dx] + C
⇒ y * log x = 2[ log x * (1-/x)  - ꭍ {(1/x) * (-1/x)} dx] + C
⇒ y * log x = 2[-log x/x + ꭍ 1/x2 dx] + C
⇒ y * log x = 2[-log x/x – 1/x] + C
⇒ y * log x = -2[1 + log x/x]/x + C
This is the required general solution of the given differential equation.

Question :8:
(1 + x2)dy + 2xy dx = cot x dx (x ≠ 0)

The given differential equation is:
(1 + x2)dy + 2xy dx = cot x dx
⇒ dy/dx + 2xy/(1 + x2) = cot x/(1 + x2)
This is in the form of dy/dx + py = Q, where p = 2x/(1 + x2) and Q = cot x/(1 + x2)
Now, I.F = eꭍp dx = eꭍ 2x/(1 + x2) dx = elog((1 + x2)) = 1 + x2
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y * (1 + x2) = ꭍ[ cot x/(1 + x2) * (1 + x2)]dx + C
⇒ y(1 + x2) = ꭍ cot x dx + C
⇒ y(1 + x2) = log |sin x| + C
This is the required general solution of the given differential equation.

Question :9:
x * dy/dx + y – x + xy cot x = 0 (x ≠ 0)

The given differential equation is:
x * dy/dx + y – x + xy cot x = 0
⇒ x * dy/dx + y(1 + x cot x) = x
⇒ dy/dx + y(1/x + cot x) = 1
This is in the form of dy/dx + py = Q, where p = 1/x + cot x and Q = 1
Now, I.F = eꭍp dx = eꭍ (1/x + cot x)dx = elog x + log(sin x) = elog(x sin x) = x sin x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y * (x sin x) = ꭍ(1 * x sin x)dx + C
⇒ y(x sin x) = x ꭍ sin x dx  - ꭍ[d(x)/dx * ꭍsin x dx]dx + C
⇒ y(x sin x) = x(-cos x)  - ꭍ{1 * ꭍ(-cos x) dx} + C
⇒ y(x sin x) = -x cos x  + sin x + C
⇒ y(x sin x) = -x cos x/(x sin x)  + sin x/(x sin x) + C/(x sin x)
⇒ y(x sin x) = -cot x  + 1/x + C/(x sin x)
This is the required general solution of the given differential equation.

Question :10:
(x + y)dy/dx = 1

The given differential equation is:
(x + y)dy/dx = 1
⇒ dy/dx = 1/(x + y)
⇒ dx/dy = x + y
⇒ dx/dy - x = y
This is in the form of dx/dy + px = Q, where p = -1 and Q = y
Now, I.F = eꭍp dy = eꭍ-dy = e-y
The solution of the given differential equation is given by the relation,
x(I.F) = ꭍ(Q * I.F.) dy + C
⇒ x e-y = ꭍ (y * e-y)dy + C
⇒ xe-y = y ꭍ e-y dy – [ ꭍ d(y)/dy * ꭍe-y dy]dy + C
⇒ xe-y = -ye-y – ꭍ (-e-y)dy + C
⇒ xe-y = -ye-y + ꭍ e-y dy + C
⇒ xe-y = -ye-y - e-y + C
⇒ x = -y - 1 + Cey
⇒ x  + y + 1 = Cey
This is the required general solution of the given differential equation.

Question :11:
y dx + (x – y2)dy = 1

The given differential equation is:
y dx + (x – y2)dy = 0
⇒ y dx = (y2 – x)dy
⇒ dx/dy = (y2 – x)/y
⇒ dx/dy = y – x/y
⇒ dx/dy + x/y = y
This is in the form of dx/dy + px = Q, where p = 1/y and Q = y
Now, I.F = eꭍp dy = eꭍ1/y dy = elog y = y
The solution of the given differential equation is given by the relation,
x(I.F) = ꭍ(Q * I.F.) dy + C
⇒ xy = ꭍ (y * y)dy + C
⇒ xy = ꭍ y2 dy + C
⇒ xy = y3/3 + C
⇒ x = y2/3 + C/y
This is the required general solution of the given differential equation.

Question :12:
(x + 3y2) * dy/dx = y (y > 0)

The given differential equation is:
(x + 3y2) * dy/dx = y
⇒ dy/dx = y/(x + 3y2)
⇒ dx/dy = (x + 3y2)/y
⇒ dx/dy = x/y + 3y
⇒ dx/dy - x/y = 3y
This is in the form of dx/dy + px = Q, where p = -1/y and Q = 3y
Now, I.F = eꭍp dy = eꭍ-dy/y = e-log y = elog (1/y) = 1/y
The solution of the given differential equation is given by the relation,
x(I.F) = ꭍ(Q * I.F.) dy + C
⇒ x * 1/y = ꭍ (3y * 1/y)dy + C
⇒ x/y = 3ꭍ dy + C
⇒ x/y = 3y + C
⇒ x = 3y2 + Cy
This is the required general solution of the given differential equation.

Question :13:
dy/dx + 2y tan x = sin x; y = 0 when x = π/3

The given differential equation is:
dy/dx + 2y tan x = sin x
This is in the form of dy/dx + py = Q, where p = 2 tan x and Q = sin x
Now, I.F = eꭍp dy = eꭍ2 tan x dx = e 2log |sec x| = elog (sec2 x) = sec2 x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y * sec2 x = ꭍ (sin x * sec2 x)dx + C
⇒ y * sec2 x = ꭍ (sec x * tan x)dx + C
⇒ y * sec2 x = sec x + C  …………1
Now, y = 0 when x = π/3
⇒ 0 * sec2 π/3 = sec π/3 + C
⇒ 0 = 2 + C
⇒ C = -2
Put the value of C in equation 1, we get
⇒ y * sec2 x = sec x – 2
⇒ y = cos x – 2 cos2 x
This is the required general solution of the given differential equation.

Question :14:
(1 + x2)dy/dx + 2xy = 1/(1 + x2); y = 0 when x = 1

The given differential equation is:
(1 + x2)dy/dx + 2xy = 1/(1 + x2)
⇒ dy/dx + 2xy/(1 + x2) = 1/(1 + x2)2
This is in the form of dy/dx + py = Q, where p = 2 tan x and Q = sin x
Now, I.F = eꭍp dy = eꭍ 2xy/(1 + x2) dx = e log (1 + x2) = (1 + x2)
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y(1 + x2) = ꭍ [1/(1 + x2) * (1 + x2)2]dx + C
⇒ y/(1 + x2) = ꭍ 1/(1 + x2) dx + C
⇒ y/(1 + x2) = tan-1 x + C  ………..1
Now, y = 0 when x = 1
⇒ 0 = tan-1 1 + C
⇒ C = -π/4
Put the value of C in equation 1, we get
⇒ y/(1 + x2) = tan-1 x - π/4
This is the required general solution of the given differential equation.

Question :15:
dy/dx - 3y cot x = sin 2x; y = 2 when x = π/2

The given differential equation is:
dy/dx - 3y cot x = sin 2x
This is in the form of dy/dx + py = Q, where p = -3 cot x and Q = sin 2x
Now, I.F = eꭍp dy = eꭍ -3 cot x dx = e3 log|sin x| = elog|1/sin3 x| = 1/sin3 x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y/sin3 x = ꭍ[sin 2x * 1/sin3 x]dx + C
⇒ y cosec3 x = 2ꭍ[cot x * cosec x]dx + C
⇒ y cosec3 x = 2 cosec x + C
⇒ y = -2/cosec2 x + C/cosec3 x
⇒ y = -2 sin2 x + C sin3 x  ………….1
Now, y = 2 when x = π/2
⇒ 2 = -2 + C
⇒ C = 4
Put the value of C in equation 1, we get
⇒ y = -2 sin2 x + 4 sin3 x
⇒ y = 4 sin3 x - 2 sin2 x
This is the required general solution of the given differential equation.

Question :16:
Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Let F (x, y) be the curve passing through the origin.
At point (x, y), the slope of the curve will be dy/dx
According to the given information:
dy/dx = x + y
⇒ dy/dx - y = x
This is a linear differential equation of the form:
dy/dx + py = Q, where p = -1 and Q = x
Now, I.F = eꭍp dy = eꭍ (-1) dx = e-x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y(I.F) = ꭍ(x * e-x) dx + C
⇒ ye-x = x ꭍ e-x dx - ꭍ[d(x)/dx * ꭍ e-x dx]dx + C
⇒ ye-x = -xe-x - ꭍ(-e-x) dx + C
⇒ ye-x = -xe-x + ꭍ e-x dx + C
⇒ ye-x = -xe-x - e-x + C
⇒ ye-x = -e-x (x + 1) + C
⇒ y = -(x + 1) + Cex
⇒ x + y + 1 = Cex  …………..1
The curve passes through the origin.
⇒ 1 = Ce0
⇒ C = 1
Put the value of C in equation 2, we get
x + y + 1 = ex
Hence, the required equation of curve passing through the origin is x + y + 1 = ex

Question :17:
Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent to the
curve at (x, y) is dy/dx
According to the given information:
dy/dx + 5 = x + y
⇒ dy/dx - y = x – 5
This is a linear differential equation of the form:
dy/dx + py = Q, where p = -1 and Q = x - 5
Now, I.F = eꭍp dy = eꭍ (-1) dx = e-x
The solution of the given differential equation is given by the relation,
y(I.F) = ꭍ(Q * I.F.) dx + C
⇒ y(I.F) = ꭍ(x – 5) * e-x dx + C
⇒ ye-x = (x – 5)ꭍ e-x dx - ꭍ[d(x – 5)/dx * ꭍ e-x dx]dx + C
⇒ ye-x = -(x – 5)e-x - ꭍ(-e-x) dx + C
⇒ ye-x = (5 - x)e-x + ꭍ e-x dx + C
⇒ ye-x = (5 - x)e-x - e-x + C
⇒ ye-x = (4 - x)e-x + C
⇒ y = 4 - x + Cex
⇒ x + y - 4 = Cex …………….1
The curve passes through the point (0, 2).
⇒ 0 + 2 - 4 = Ce0
⇒ C = -2
Put the value of C in equation 2, we get
x + y - 4 = -2ex
⇒ y = 4 – x - 2ex
This is the required equation of the curve.

Question :18:
The integrating factor of the differential equation is x * dy/dx – y = 2x2                                                                        A. e–x                        B. e–y                     C. 1/x                                    D. x

The given differential equation is:
x * dy/dx – y = 2x2
⇒ dy/dx – y/x = 2x
This is a linear differential equation of the form:
This is a linear differential equation of the form:
dy/dx + py = Q, where p = -1/x and Q = 2x
Now, I.F = eꭍp dy = eꭍ (-1/x) dx = e-log(x) = elog(x-1) = x-1 = 1/x
Hence, the correct answer is option C.

Question :19:
The integrating factor of the differential equation.
(1 - y2) * dy/dx + yx = ay  (-1< y < 1)

1. 1/(y2 - 1) B. 1/√(y2 - 1)         C. 1/(1 - y2)                            D. 1/√(1 - y2)