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

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) $$\frac{d y}{d x}$$ = x2y + y
Solution:
$$\frac{d y}{d x}$$ = x2y + y
∴ $$\frac{d y}{d x}$$ = y(x2 + 1)
∴ $$\frac{1}{y}$$ dy = (x2 + 1) dx
Integrating, we get
∫$$\frac{1}{y}$$ dy = ∫(x2 + 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) (x2 – yx2) dy + (y2 + xy2) dx = 0
Solution:
(x2 – yx2) dy + (y2 + xy2) dx = 0
∴ x2(1 – y) dy + y2(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:
$$y^{3}-\frac{d y}{d x}=x \frac{d y}{d x}$$

∴ 2y2 log |x + 1| = 2cy2 – 1 is the required solution.

Question 2.
For each of the following differential equations find the particular solution:
(i) (x – y2x) dx – (y + x2y) dy = 0, when x = 2, y = 0.
Solution:
(x – y2x) dx – (y + x2y) dy = 0
∴ x(1 – y2) dx – y(1 + x2) dy = 0

∴ the general solution is
log |1 + x2| + log |1 – y2| = log c, where c1 = log c
∴ log |(1 + x2)(1 – y2) | = log c
∴ (1 + x2)(1 – y2) = c
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x2)(1 – y2) = 5.

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

∴ log |2 + ey| = log |c(x + 1)|
∴ 2 + ey = c(x + 1)
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e0 = c(1 + 1)
∴ 3 = 2c
∴ c = $$\frac{3}{2}$$
∴ the particular solution is
2 + ey = $$\frac{3}{2}$$(x + 1)
∴ 4 + 2ey = 3x + 3
∴ 3x – 2ey – 1 = 0

(iii) y(1 + log x) $$\frac{d x}{d y}$$ – x log x = 0, when x = e, y = e2.
Solution:

∴ from (1), the general solution is
log |x log x| – log |y| = log c, where c1 = 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 = e2, when x = e
e log e = ce2
1 = ce ……[∵ log e = 1]
c = $$\frac{1}{e}$$
∴ the particular solution is x log x = ($$\frac{1}{e}$$) y
∴ y = ex 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