Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Miscellaneous Exercise 6 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Miscellaneous Exercise 6

(I) Choose the correct option from the given alternatives:

Question 1.

The order and degree of the differential equation \(\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d^{2} y}{d x^{2}}\right)^{\frac{3}{2}}\) are respectively……..

(a) 2, 1

(b) 1, 2

(c) 3, 2

(d) 2, 3

Answer:

(d) 2, 3

Question 2.

The differential equation of y = c^{2} + \(\frac{c}{x}\) is…….

(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)

(b) \(\frac{d y}{d x^{2}}+x \frac{d y}{d x}+y=0\)

(c) \(x^{3}\left(\frac{d y}{d x}\right)^{2}+x \frac{d y}{d x}=y\)

(d) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)

Answer:

(a) \(x^{4}\left(\frac{d y}{d x}\right)^{2}-x \frac{d y}{d x}=y\)

Question 3.

x^{2} + y^{2} = a^{2} is a solution of ………

(a) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}-y=0\)

(b) \(y=x \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}+a^{2} y\)

(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)

(d) \(\frac{d^{2} y}{d x^{2}}=(x+1) \frac{d y}{d x}\)

Answer:

(c) \(y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\)

Question 4.

The differential equation of all circles having their centres on the line y = 5 and touching the X-axis is

(a) \(y^{2}\left(1+\frac{d y}{d x}\right)=25\)

(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)

(c) \((y-5)^{2}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)

(d) \((y-5)^{2}\left[1-\left(\frac{d y}{d x}\right)^{2}\right]=25\)

Answer:

(b) \((y-5)^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right]=25\)

Question 5.

The differential equation y \(\frac{d y}{d x}\) + x = 0 represents family of ………

(a) circles

(b) parabolas

(c) ellipses

(d) hyperbolas

Answer:

(a) circles

Hint:

y \(\frac{d y}{d x}\) + x = 0

∴ ∫y dy + ∫x dx = c

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

∴ x^{2} + y^{2} = 2c which is a circle.

Question 6.

The solution of \(\frac{1}{x} \cdot \frac{d y}{d x}=\tan ^{-1} x\) is……

(a) \(\frac{x^{2} \tan ^{-1} x}{2}+c=0\)

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

(c) x – tan^{-1}x = c

(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)

Answer:

(d) \(y=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c\)

Question 7.

The solution of (x + y)^{2} \(\frac{d y}{d x}\) = 1 is…….

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

(b) y tan^{-1}(\(\frac{x}{y}\)) = c

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

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

Answer:

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

Question 8.

The Solution of \(\frac{d y}{d x}=\frac{y+\sqrt{x^{2}-y^{2}}}{2}\) is……

(a) sin^{-1}(\(\frac{y}{x}\)) = 2 log |x| + c

(b) sin^{-1}(\(\frac{y}{x}\)) = log |x| + c

(c) sin(\(\frac{x}{y}\)) = log |x| + c

(d) sin(\(\frac{y}{x}\)) = log |y| + c

Answer:

(b) sin^{-1}(\(\frac{y}{x}\)) = log |x| + c

Question 9.

The solution of \(\frac{d y}{d x}\) + y = cos x – sin x is……

(a) y e^{x} = cos x + c

(b) y e^{x} + e^{x} cos x = c

(c) y e^{x} = e^{x} cos x + c

(d) y^{2} e^{x} = e^{x} cos x + c

Answer:

(c) y e^{x} = e^{x} cos x + c

Hint:

\(\frac{d y}{d x}\) + y = cos x – sin x

I.F. = \(e^{\int 1 d x}=e^{x}\)

∴ the solution is y . e^{x} = ∫(cos x – sin x) e^{x} + c

∴ y . e^{x} = e^{x} cos x + c

Question 10.

The integrating factor of linear differential equation x \(\frac{d y}{d x}\) + 2y = x^{2} log x is……..

(a) \(\frac{1}{x}\)

(b) k

(c) \(\frac{1}{n^{2}}\)

(d) x^{2}

Answer:

(d) x^{2}

Hint:

I.F. = \(e^{\int \frac{2}{x} d x}\)

= e^{2 log x}

= x^{2}

Question 11.

The solution of the differential equation \(\frac{d y}{d x}\) = sec x – y tan x is…….

(a) y sec x + tan x = c

(b) y sec x = tan x + c

(c) sec x + y tan x = c

(d) sec x = y tan x + c

Answer:

(b) y sec x = tan x + c

Hint:

\(\frac{d y}{d x}\) = sec x – y tan x

∴ \(\frac{d y}{d x}\) + y tan x = sec x

I.F. = \(e^{\int \tan x d x}=e^{\log \sec x}\) = sec x

∴ the solution is

y . sec x = ∫sec x . sec x dx + c

∴ y sec x = tan x + c

Question 12.

The particular solution of \(\frac{d y}{d x}=x e^{y-x}\), when x = y = 0 is……

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

(b) e^{x+y} = x + 1

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

(d) e^{y-x} = x – 1

Answer:

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

Question 13.

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is a solution of……..

(a) \(\frac{d^{2} y}{d x^{2}}+y x+\left(\frac{d y}{d x}\right)^{2}=0\)

(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)

(c) \(y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}+y=0\)

(d) \(x y \frac{d y}{d x}+y \frac{d^{2} y}{d x^{2}}=0\)

Answer:

(b) \(x y \frac{d^{2} y}{d x^{2}}+2\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0\)

Question 14.

The decay rate of certain substances is directly proportional to the amount present at that instant. Initially, there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is……

(a) 5\(\frac{2}{3}\) grams

(b) 5\(\frac{1}{3}\) grams

(c) 5.1 grams

(d) 5 grams

Answer:

(b) 5\(\frac{1}{3}\) grams

Question 15.

If the surrounding air is kept at 20°C and the body cools from 80°C to 70°C in 5 minutes, the temperature of the body after 15 minutes will be…..

(a) 51.7°C

(b) 54.7°C

(c) 52.7°C

(d) 50.7°C

Answer:

(b) 54.7°C

(II) Solve the following:

Question 1.

Determine the order and degree of the following differential equations:

(i) \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)

Solution:

The given D.E. is \(\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+y=x^{3}\)

This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.

∴ the given D.E. is of order 2 and degree 1.

(ii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)

Solution:

The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}=\sqrt[5]{1+\frac{d y}{d x}}\)

\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2 \times 5}=1+\frac{d y}{d x}\)

\(\left(\frac{d^{3} y}{d x^{3}}\right)^{10}=1+\frac{d y}{d x}\)

This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 10.

∴ the given D.E. is of order 3 and degree 10.

(iii) \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)

Solution:

The given D.E. is \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)

On cubing both sides, we get

\(1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\)

This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 3.

∴ the given D.E. is of order 2 and degree 3.

(iv) \(\frac{d y}{d x}=3 y+\sqrt[4]{1+5\left(\frac{d y}{d x}\right)^{2}}\)

Solution:

The given D.E. is

This D.E. has the highest order derivative \(\frac{d y}{d x}\) with power 4.

∴ the given D.E. is of order 1 and degree 4.

(v) \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)

Solution:

The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\sin \left(\frac{d y}{d x}\right)=0\)

This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\).

∴ order = 4

Since this D.E. cannot be expressed as a polynomial in differential coefficient, the degree is not defined.

Question 2.

In each of the following examples verify that the given function is a solution of the differential equation.

(i) \(x^{2}+y^{2}=r^{2} ; x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)

Solution:

x^{2} + y^{2} = r^{2} ……. (1)

Differentiating both sides w.r.t. x, we get

Hence, x^{2} + y^{2} = r^{2} is a solution of the D.E.

\(x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=y\)

(ii) y = e^{ax} sin bx; \(\frac{d^{2} y}{d x^{2}}-2 a \frac{d y}{d x}+\left(a^{2}+b^{2}\right) y=0\)

Solution:

(iii) y = 3 cos(log x) + 4 sin(log x); \(x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\)

Solution:

y = 3 cos(log x) + 4 sin (log x) …… (1)

Differentiating both sides w.r.t. x, we get

(iv) xy = ae^{x} + be^{-x} + x^{2}; \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+x^{2}=x y+2\)

Solution:

(v) x^{2} = 2y^{2} log y, x^{2} + y^{2} = xy \(\frac{d x}{d y}\)

Solution:

x^{2} = 2y^{2} log y ……(1)

Differentiating both sides w.r.t. y, we get

∴ x^{2} + y^{2} = xy \(\frac{d x}{d y}\)

Hence, x^{2} = 2y^{2} log y is a solution of the D.E.

x^{2} + y^{2} = xy \(\frac{d x}{d y}\)

Question 3.

Obtain the differential equation by eliminating the arbitrary constants from the following equations:

(i) y^{2} = a(b – x)(b + x)

Solution:

y^{2} = a(b – x)(b + x) = a(b^{2} – x^{2})

Differentiating both sides w.r.t. x, we get

2y \(\frac{d y}{d x}\) = a(0 – 2x) = -2ax

∴ y \(\frac{d y}{d x}\) = -ax …….(1)

Differentiating again w.r.t. x, we get

This is the required D.E.

(ii) y = a sin(x + b)

Solution:

y = a sin(x + b)

This is the required D.E.

(iii) (y – a)^{2} = b(x + 4)

Solution:

(y – a)^{2} = b(x + 4) …….(1)

Differentiating both sides w.r.t. x, we get

\(2(y-a) \cdot \frac{d}{d x}(y-a)=b \frac{d}{d x}(x+4)\)

(iv) y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)

Solution:

y = \(\sqrt{a \cos (\log x)+b \sin (\log x)}\)

∴ y^{2} = a cos (log x) + b sin (log x) …….(1)

Differentiating both sides w.r.t. x, we get

(v) y = Ae^{3x+1} + Be^{-3x+1}

Solution:

y = Ae^{3x+1} + Be^{-3x+1} …… (1)

Differentiating twice w.r.t. x, we get

This is the required D.E.

Question 4.

Form the differential equation of:

(i) all circles which pass through the origin and whose centres lie on X-axis.

Solution:

Let C (h, 0) be the centre of the circle which pass through the origin. Then radius of the circle is h.

∴ equation of the circle is (x – h)^{2} + (y – 0)^{2} = h^{2}

∴ x^{2} – 2hx + h^{2} + y^{2} = h^{2}

∴ x^{2} + y^{2} = 2hx ……..(1)

Differentiating both sides w.r.t. x, we get

2x + 2y \(\frac{d y}{d x}\) = 2h

Substituting the value of 2h in equation (1), we get

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

∴ x^{2} + y^{2} = 2x^{2} + 2xy \(\frac{d y}{d x}\)

∴ 2xy \(\frac{d y}{d x}\) + x^{2} – y^{2} = 0

This is the required D.E.

(ii) all parabolas which have 4b as latus rectum and whose axis is parallel to Y-axis.

Solution:

Let A(h, k) be the vertex of the parabola which has 4b as latus rectum and whose axis is parallel to the Y-axis.

Then equation of the parabola is

(x – h)^{2} = 4b(y – k) ……. (1)

where h and k are arbitrary constants.

Differentiating both sides of (1) w.r.t. x, we get

2(x – h). \(\frac{d}{d x}\)(x – h) = 4b . \(\frac{d}{d x}\)(y – k)

∴ 2(x – h) x (1 – 0) = 4b(\(\frac{d y}{d x}\) – 0)

∴ (x – h) = 2b \(\frac{d y}{d x}\)

Differentiating again w.r.t. x, we get

1 – 0 = 2b \(\frac{d^{2} y}{d x^{2}}\)

∴ 2b \(\frac{d^{2} y}{d x^{2}}\) – 1 = 0

This is the required D.E.

(iii) an ellipse whose major axis is twice its minor axis.

Solution:

Let 2a and 2b be lengths of the major axis and minor axis of the ellipse.

Then 2a = 2(2b)

∴ a = 2b

∴ equation of the ellipse is

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

∴ \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)

∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)

∴ x^{2} + 4y^{2} = 4b^{2}

Differentiating w.r.t. x, we get

2x + 4 × 2y \(\frac{d y}{d x}\) = 0

∴ x + 4y \(\frac{d y}{d x}\) = 0

This is the required D.E.

(iv) all the lines which are normal to the line 3x + 2y + 7 = 0.

Solution:

Slope of the line 3x – 2y + 7 = 0 is \(\frac{-3}{-2}=\frac{3}{2}\).

∴ slope of normal to this line is \(-\frac{2}{3}\)

Then the equation of the normal is

y = \(-\frac{2}{3}\)x + k, where k is an arbitrary constant.

Differentiating w.r.t. x, we get

\(\frac{d y}{d x}=-\frac{2}{3} \times 1+0\)

∴ 3\(\frac{d y}{d x}\) + 2 = 0

This is the required D.E.

(v) the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\).

Solution:

The equation of the hyperbola is \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=k\)

i.e., \(\frac{x^{2}}{16 k}-\frac{y^{2}}{36 k}=1\)

Comparing this equation with \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), we get

a^{2} = 16k, b^{2} = 36k

∴ a = 4√k, b = 6√k

∴ l(transverse axis) = 2a = 8√k

and l(conjugate axis) = 2b = 12√k

Let 2A and 2B be the lengths of the transverse and conjugate axes of the required hyperbola.

Then according to the given condition

2A = a = 4√k and 2B = b = 6√k

∴ A = 2√k and B = 3√k

∴ equation of the required hyperbola is

\(\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1\)

i.e., \(\frac{x^{2}}{4 k}-\frac{y^{2}}{9 k}=1\)

∴ 9x^{2} – 4y^{2} = 36k, where k is an arbitrary constant.

Differentiating w.r.t. x, we get

9 × 2x – 4 × 2y \(\frac{d y}{d x}\) = 0

∴ 9x – 4y \(\frac{d y}{d x}\) = 0

This is the required D.E.

Question 5.

Solve the following differential equations:

(i) log(\(\frac{d y}{d x}\)) = 2x + 3y

Solution:

(ii) \(\frac{d y}{d x}\) = x^{2}y + y

Solution:

(iii) \(\frac{d y}{d x}=\frac{2 y-x}{2 y+x}\)

Solution:

(iv) x dy = (x + y + 1) dx

Solution:

(v) \(\frac{d y}{d x}\) + y cot x = x^{2} cot x + 2x

Solution:

\(\frac{d y}{d x}\) + y cot x = x cot x + 2x ……..(1)

This is the linear differential equation of the form

\(\frac{d y}{d x}\) + Py = Q, where P = cot x and Q = x^{2} cot x + 2x

∴ I.F. = \(e^{\int P d x}\)

= \(e^{\int \cot x d x}\)

= \(e^{\log (\sin x)}\)

= sin x

∴ the solution of (1) is given by

y(I.F.) = ∫Q . (I.F.) dx + c

∴ y sin x = ∫(x^{2} cot x + 2x) sin x dx + c

∴ y sinx = ∫(x^{2} cot x . sin x + 2x sin x) dx + c

∴ y sinx = ∫x^{2} cos x dx + 2∫x sin x dx + c

∴ y sinx = x^{2} ∫cos x dx – ∫[\(\frac{d}{d x}\left(x^{2}\right)\) ∫cos x dx] dx + 2∫x sin x dx + c

∴ y sin x = x^{2} (sin x) – ∫2x(sin x) dx + 2∫x sin x dx + c

∴ y sin x = x^{2} sin x – 2∫x sin x dx + 2∫x sin x dx + c

∴ y sin x = x^{2} sin x + c

∴ y = x^{2} + c cosec x

This is the general solution.

(vi) y log y = (log y^{2} – x) \(\frac{d y}{d x}\)

Solution:

(vii) 4 \(\frac{d x}{d y}\) + 8x = 5e^{-3y}

Solution:

Question 6.

Find the particular solution of the following differential equations:

(i) y(1 + log x) = (log xx) \(\frac{d y}{d x}\), when y(e) = e^{2}

Solution:

(ii) (x + 2y^{2}) \(\frac{d y}{d x}\) = y, when x = 2, y = 1

Solution:

This is the general solution.

When x = 2, y = 1, we have

2 = 2(1)^{2} + c(1)

∴ c = 0

∴ the particular solution is x = 2y^{2}.

(iii) \(\frac{d y}{d x}\) – 3y cot x = sin 2x, when y(\(\frac{\pi}{2}\)) = 2

Solution:

\(\frac{d y}{d x}\) – 3y cot x = sin 2x

\(\frac{d y}{d x}\) = (3 cot x) y = sin 2x ……..(1)

This is the linear differential equation of the form

(iv) (x + y) dy + (x – y) dx = 0; when x = 1 = y

Solution:

(v) \(2 e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0\), when y(0) = 1

Solution:

Question 7.

Show that the general solution of defferential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0\) is given by (x + y + 1) = c(1 – x – y – 2xy).

Solution:

Question 8.

The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.

Solution:

Let P(x, y) be a point on the curve y = f(x).

Then slope of the normal to the curve is \(-\frac{1}{\left(\frac{d y}{d x}\right)}\)

∴ equation of the normal is

This is the general equation of the curve.

Since, the required curve passed through the point (2, 3), we get

2^{2} + 3^{2} = 4(2) + c

∴ c = 5

∴ equation of the required curve is x^{2} + y^{2} = 4x + 5.

Question 9.

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Solution:

Let r be the radius and V be the volume of the spherical balloon at any time t.

Then the rate of change in volume of the spherical balloon is \(\frac{d V}{d t}\) which is a constant.

Hence, the radius of the spherical balloon after t seconds is \((63 t+27)^{\frac{1}{3}}\) units.

Question 10.

A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning are ₹ 10 lakhs and they dwindle down to ₹ 10,000 after 2 years, show that the person will be bankrupt in 2\(\frac{2}{9}\) years from the start.

Solution:

Let x be the assets of the presort at time t years.

Then the rate of reduction is \(\frac{d x}{d t}\) which is proportional to √x.

∴ \(\frac{d x}{d t}\) ∝ √x

∴ \(\frac{d x}{d t}\) = -k√x, where k > 0

∴ \(\frac{d x}{\sqrt{x}}\) = -k dt

Integrating both sides, we get

\(\int x^{-\frac{1}{2}} d x\) = -k∫dt

∴ \(\frac{x^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\) = -kt + c

∴ 2√x = -kt + c

At the beginning, i.e. at t = 0, x = 10,00,000

2√10,00,000 = -k(0) + c

∴ c = 2000

∴ 2√x = -kt + 2000 ……..(1)

Also, when t = 2, x = 10,000

∴ 2√10000 = -k × 2 + 2000

∴ 2k = 1800

∴ k = 900

∴ (1) becomes,

∴ 2√x = -900t + 2000

When the person will be bankrupt, x = 0

∴ 0 = -900t + 2000

∴ 900t = 2000

∴ t = \(\frac{20}{9}=2 \frac{2}{9}\)

Hence, the person will be bankrupt in \(2 \frac{2}{9}\) years.