# Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4

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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4

Question 1.
Show that each of the following expressions is a solution of the corresponding given differential equation.

(i) y = 2x2 ; xy’ = 2y
Solution:
v = 2x2 …(1)
Differential equation: xy’ = 2y
Differentiate with respect to ‘x’ On simplifying, 2y = xy’
∴ (1) is solution of the given differential equation.

(ii) y = aex + be-x ; y” – y = 0
Solution:
y = aex + be-x …(1) Differential equation: y” – y = 0
Differentiate with respect to ‘x’
y’ = aex – be-x
Again differentiate with respect to ‘x’
y” = aex + be-x
y” = y ⇒ y” – y = 0
∴ (1) is the solution of the given differential equation. Question 2.
Find value of m so that the function y = emx is a solution of the given differential equation.
(i) y + 2y = 0
Solution:
Given solution y = emx
Differentiate with respect to ‘x’ (ii) y” – 5y’ + 6y = 0 Given differential equation is y” – 5y’ + 6y = 0
Substitute (1), (2) and (3) in this Question 3.
The slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through (2, 5). Find the equation of the curve.
Solution:
Slope of the tangent is the reciprocal of four times the ordinate Question 4. Solution:  Question 5.
Show that y = ax + $$\frac{b}{x}$$, x ≠ 0, is a solution of the differential equation x2y”+ xy’ – y = 0.
Solution: Here ‘a’ and ‘b’ are arbitrary constants Differentiate with respect to ‘x’
xy’ + y . 1 = a (2x) = 2ax ……….. (2)
Differentiate again with respect to ‘x’ .
xy” + y’ . 1 + y = 2a ⇒ xy” + 2y’ = 2a …….. (3)
Substitute (3) in (2)
xy’ + y = (xy” + 2y’)x
xy’ + y = x2y” + 2xy’ ⇒ x2y” + xy’ – y = 0
Hence proved.

Question 6.
Show that y = ae-3x + b, where a and b are arbitrary constants, is a solution of the differential equation Solution: Question 7.
Show that the differential equation representing the family of curves where a is a positive parameter, is Solution:  Question 8.
Show that, y = a cos bx is a solution of the differential equation .
Solution:
y = a cos bx …(1) (a is an arbitrary constant)
Differentiating with respect to ‘x’ Again, differentiating with respect to ‘x’ Hence proved

### Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.4 Additional Problems

Question 1.
Verify that the function y = a cos x + b sin x is a solution of the differential equation cos $$\frac{d y}{d x}$$ + y sin x = b. dx
Solution:
The given function is y = a cos x + b sin x
Differentiating both sides with respect to x, we have Putting values of $$\frac{d y}{d x}$$ and y in the given differential equation, we have
L.H.S. = cos x (- a sin x + b cos x) + {a cos x + b sin x) sin x
= – a sin x cos x + b cos2 x + a sin x cos x + b sin2 x = b (cos2 x + sin2 x)
= b × 1 = b = R.H.S
Thus, y = a cos x + b sin x is a solution of differential equation  Question 2.
Verify that the function y = 4 sin 3x is a solution of the differential equation Solution:
The given function is y = 4 sin 3x
Differentiating both sides with respect to x, we have Again, differentiating both sides with respect to x, we have Putting values of $$\frac{d^{2} y}{d x^{2}}$$ and y in the given differential equation, we have
L.H.S. = – 36 sin 3x + 9 (4 sin 3x) = – 36 sin 3x + 36 sin 3x = 0 = R.H.S.
Thus, y = 4 sin 3x is a solution of differential equation Question 3.
Verify that the function y = ax2 + bx + c is a solution of the differential equation .
Solution:
The given function is y = ax2 + bx + c
Differentiating both sides with respect to x, we have Again differentiating both sides with respect to x, we have Which is the given differential equation.
Thus, y = ax2 + bx + c is a solution of differential equation $$\frac{d^{2} y}{d x^{2}}$$= 2a. Question 4.
Verify that the function y = e-3x is a solution of the differential equation Solution:
The given function is y = e-3x
Differentiating both sides with respect to x, we have Again, differentiating both sides with respect to x, we have 