Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 6 Differential Equations Ex 6.5 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.5
Solution & Step-by-Step Answer:
…….(1) This is the linear differential equation of the form + P. y = Q, where P = and Q = x3 – 3 This is the general solution.
(ii) cos2x. + y = tan x
Solution:
(iii) (x + 2y3) = y
Solution:
(iv) + y. sec x = tan x
Solution:
+ y sec x = tan x
∴ + (sec x). y = tan x ……..(1)
This is the linear differential equation of the form
+ P. y = Q, where P = sec x and Q = tan x
∴ I.F. =
=
=
= sec x + tan x
∴ the solution of (1) is given by
y (I.F.) = ∫Q. (I.F.) dx + c
∴ y(sec x + tan x) = ∫tan x (sec x + tan x) dx + c
∴ (sec x + tan x). y = ∫(sec x tan x + tan2x) dx + c
∴ (sec x + tan x). y = ∫(sec x tan x + sec2x – 1) dx + c
∴ (sec x + tan x). y = sec x + tan x – x + c
∴ y(sec x + tan x) = sec x + tan x – x + c
This is the general solution.
(v) x + 2y = x2. log x
Solution:
(vi) (x + y) = 1
Solution:
(x + y) = 1
∴ = x + y
∴ – x = y
∴ + (-1) x = y ……….(1)
This is the linear differential equation of the form
This is the general solution.
(vii) (x + a) – 3y = (x + a)5
Solution:
(viii) dr + (2r cot θ + sin 2θ) dθ = 0
Solution:
dr + (2r cot θ + sin 2θ) dθ = 0
∴ + (2r cot θ + sin 2θ) = 0
∴ + (2 cot θ)r = -sin 2θ ………(1)
This is the linear differential equation of the form dr
+ P. r = Q, where P = 2 cot θ and Q = -sin 2θ
This is the general solution.
(ix) y dx + (x – y2) dy = 0
Solution:
y dx + (x – y2) dy = 0
∴ y dx = -(x – y2) dy
∴
∴ ………(1)
This is the linear differential equation of the form
This is the general solution.
(x)
Solution:
(xi)
Solution:
Solution & Step-by-Step Answer:
Let A(x, y) be the point on the curve y = f(x). Then slope of the tangent to the curve at the point A is . According to the given condition, = x + 3y – 1 ∴ – 3y = x – 1 ………(1) This is the linear differential equation of the form This is the general equation of the curve. But the required curve is passing through the origin (0, 0). ∴ by putting x = 0 and y = 0 in (2), we get 0 = 2 + c ∴ c = -2 ∴ from (2), the equation of the required curve is 3(x + 3y) = 2 – 2e3x i.e. 3(x + 3y) = 2 (1 – e3x).
Solution & Step-by-Step Answer:
Let A(x, y) be the point on the curve y = f(x). Then the slope of the tangent to the curve at point A is . According to the given condition ∴ y dy = x dx Integrating both sides, we get ∫y dy= ∫x dx ∴ ∴ 9y2 = -4x2 + 18c1 ∴ 4x2 + 9y2 = c where c = 18c1 This is the general equation of the curve. But the required curve is passing through the point . ∴ by putting x = and y = √2 in (1), we get ∴ 18 + 18 = c ∴ c = 36 ∴ from (1), the equation of the required curve is 4x2 + 9y2 = 36.
Solution & Step-by-Step Answer:
Let A(x, y) be any point on the curve. Then slope of the tangent to the curve at the point A is . According to the given condition x + y = + 5 ∴ – y = x – 5 ………(1) This is the linear differential equation of the form + P. y = Q, where P = -1 and Q = x – 5 ∴ I.F. = ∴ the solution of (1) is given by This is the general equation of the curve. But the required curve is passing through the point (0, 2). ∴ by putting x = 0, y = 2 in (2), we get 2 = 4 – 0 + c ∴ c = -2 ∴ from (2), the equation of the required curve is y = 4 – x – 2ex.
Solution & Step-by-Step Answer:
Let A(x, y) be the point on the curve y = f(x). Then slope of the tangent to the curve at the point A is . According to the given condition = x + xy ∴ – xy = x ……….. (1) This is the linear differential equation of the form + Py = Q, where P = -x and Q = x ∴ I.F. = ∴ the solution of (1) is given by y. (I.F.) = ∫Q. (I.F.) dx + c This is the general equation of the curve. But the required curve is passing through the point (0, 1). ∴ by putting x = 0 and y = 1 in (2), we get 1 + 1 = c ∴ c = 2 ∴ from (2), the equation of the required curve is 1 + y = .
