Balbharati Maharashtra State BoardStd 12 Commerce Statistics Part 1 Digest PdfChapter 8 Differential Equation and Applications Ex 8.2 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.2
Question 1
Maharashtra Board Solution
Obtain the differential equation by eliminating arbitrary constants from the following equations: (i) y = Ae3x + Be-3x
Solution & Step-by-Step Answer:
y = Ae3x + Be-3x ……(1) Differentiating twice w.r.t. x, we get This is the required D.E.
(ii) y =
Solution:
y =
∴ xy = c2x + c1
Differentiating w.r.t. x, we get
(iii) y = (c1+ c2x) ex
Solution:
y = (c1+ c2x) ex
This is the required D.E.
(iv) y = c1e3x+ c2e2x
Solution:
This is the required D.E.
(v) y2= (x + c)3
Solution:
y2= (x + c)3
Differentiating w.r.t. x, we get
This is the required D.E.
Question 2
Maharashtra Board Solution
Find the differential equation by eliminating arbitrary constant from the relation x2 + y2 = 2ax.
Solution & Step-by-Step Answer:
x2 + y2 = 2ax Differentiating both sides w.r.t. x, we get 2x + 2y = 2a Substituting value of 2a in equation (1), we get x2 + y2 = [2x + 2y ]x = 2x2 + 2xy ∴ 2xy = y2 – x2 is the required D.E.
Question 3
Maharashtra Board Solution
Form the differential equation by eliminating arbitrary constants from the relation bx + ay = ab.
Solution & Step-by-Step Answer:
bx + ay = ab ∴ ay = -bx + ab ∴ y = Differentiating w.r.t. x, we get Differentiating again w.r.t. x, we get = 0 is the required D.E.
Question 4
Maharashtra Board Solution
Find the differential equation whose general solution is x3 + y3 = 35ax.
Solution & Step-by-Step Answer:
Question 5. Form the differential equation from the relation x2 + 4y2 = 4b2. Sol ution: x2 + 4y2 = 4b2 Differentiating w.r.t. x, we get 2x + 4(2y) = 0 i.e. x + 4y = 0 is the required D.E. |