Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.3 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.3

Question 1.

Solve the following differential equations:

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

Solution:

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

∴ \(\frac{d y}{d x}\) = y(x^{2} + 1)

∴ \(\frac{1}{y}\) dy = (x^{2} + 1) dx

Integrating, we get

∫\(\frac{1}{y}\) dy = ∫(x^{2} + 1) dx

∴ log |y|= \(\frac{x^{3}}{3}\) + x + c

This is the general solution.

(ii) \(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)\)

Solution:

\(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)\)

This is the general solution.

(iii) (x^{2} – yx^{2}) dy + (y^{2} + xy^{2}) dx = 0

Solution:

(x^{2} – yx^{2}) dy + (y^{2} + xy^{2}) dx = 0

∴ x^{2}(1 – y) dy + y^{2}(1 + x) dx = 0

∴ \(\frac{1-y}{y^{2}} d y+\frac{1+x}{x^{2}} d x=0\)

Integrating, we get

This is the general solution.

(iv) \(y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}\)

Solution:

\(y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}\)

∴ 2y^{2} log |x + 1| = 2cy^{2} – 1 is the required solution.

Question 2.

For each of the following differential equations find the particular solution:

(i) (x – y^{2}x) dx – (y + x^{2}y) dy = 0, when x = 2, y = 0.

Solution:

(x – y^{2}x) dx – (y + x^{2}y) dy = 0

∴ x(1 – y^{2}) dx – y(1 + x^{2}) dy = 0

∴ the general solution is

log |1 + x^{2}| + log |1 – y^{2}| = log c, where c_{1} = log c

∴ log |(1 + x^{2})(1 – y^{2}) | = log c

∴ (1 + x^{2})(1 – y^{2}) = c

When x = 2, y = 0, we have

(1 + 4)(1 – 0) = c

∴ c = 5

∴ the particular solution is (1 + x^{2})(1 – y^{2}) = 5.

(ii) (x + 1) \(\frac{d y}{d x}\) -1 = 2e^{-y}, when y = 0, x = 1.

Solution:

∴ log |2 + e^{y}| = log |c(x + 1)|

∴ 2 + e^{y} = c(x + 1)

This is the general solution.

Now, y = 0, when x = 1

∴ 2 + e^{0} = c(1 + 1)

∴ 3 = 2c

∴ c = \(\frac{3}{2}\)

∴ the particular solution is

2 + e^{y} = \(\frac{3}{2}\)(x + 1)

∴ 4 + 2e^{y} = 3x + 3

∴ 3x – 2e^{y} – 1 = 0

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, when x = e, y = e^{2}.

Solution:

∴ from (1), the general solution is

log |x log x| – log |y| = log c, where c_{1} = log c

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

∴ \(\frac{x \log x}{y}\) = c

∴ x log x = cy

This is the general solution.

Now, y = e^{2}, when x = e

e log e = ce^{2}

1 = ce ……[∵ log e = 1]

c = \(\frac{1}{e}\)

∴ the particular solution is x log x = (\(\frac{1}{e}\)) y

∴ y = e^{x} log x

(iv) \(\frac{d y}{d x}\) = 4x + y + 1, when y = 1, x = 0.

Solution:

\(\frac{d y}{d x}\) = 4x + y + 1

Put 4x + y + 1 = v

∴ log |v + 4| = x + c

∴ log |4x + y + 1 + 4| = x + c

i.e. log |4x + y + 5| = x + c

This is the general solution.

Now, y = 1 when x = 0

∴ log|0 + 1 + 5| = 0 + c,

i.e. c = log 6

∴ the particular solution is

log |4x + y + 5| = x + log 6

∴ \(\log \left|\frac{4 x+y+5}{6}\right|\) = x