Differential Equation and Applications Class 12 Commerce Maths 1 Chapter 8 Exercise 8.2 Answers Maharashtra Board
Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.2 Questions and Answers.
Std 12 Maths 1 Exercise 8.2 Solutions Commerce Maths
Question 1.
Obtain the differential equation by eliminating arbitrary constants from the following equations:
(i) y = Ae
3x
+ Be
-3x
Solution:
y = Ae
3x
+ Be
-3x
……(1)
Differentiating twice w.r.t. x, we get
This is the required D.E.
(ii) y = \(c_{2}+\frac{c_{1}}{x}\)
Solution:
y = \(c_{2}+\frac{c_{1}}{x}\)
∴ xy = c
2
x + c
1
Differentiating w.r.t. x, we get
(iii) y = (c
1
+ c
2
x) e
x
Solution:
y = (c
1
+ c
2
x) e
x
This is the required D.E.
(iv) y = c
1
e
3x
+ c
2
e
2x
Solution:
This is the required D.E.
(v) y
2
= (x + c)
3
Solution:
y
2
= (x + c)
3
Differentiating w.r.t. x, we get
This is the required D.E.
Question 2.
Find the differential equation by eliminating arbitrary constant from the relation x
2
+ y
2
= 2ax.
Solution:
x
2
+ y
2
= 2ax
Differentiating both sides w.r.t. x, we get
2x + 2y\(\frac{d y}{d x}\) = 2a
Substituting value of 2a in equation (1), we get
x
2
+ y
2
= [2x + 2y \(\frac{d y}{d x}\)]x = 2x
2
+ 2xy \(\frac{d y}{d x}\)
∴ 2xy \(\frac{d y}{d x}\) = y
2
– x
2
is the required D.E.
Question 3.
Form the differential equation by eliminating arbitrary constants from the relation bx + ay = ab.
Solution:
bx + ay = ab
∴ ay = -bx + ab
∴ y = \(-\frac{b}{a} x+b\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=-\frac{b}{a} \times 1+0=-\frac{b}{a}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = 0 is the required D.E.
Question 4.
Find the differential equation whose general solution is x
3
+ y
3
= 35ax.
Solution:
Question 5.
Form the differential equation from the relation x
2
+ 4y
2
= 4b
2
.
Sol ution:
x
2
+ 4y
2
= 4b
2
Differentiating w.r.t. x, we get
2x + 4(2y\(\frac{d y}{d x}\)) = 0
i.e. x + 4y\(\frac{d y}{d x}\) = 0 is the required D.E.