In each of Exercises 1 to 10, verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

Question 1. y = e^x + 1 : y” – y = 0

Solution:

Given: y = e^x + 1, differentiating both sides w.r.t.x., we have:

[Tex]\frac{dy}{dx}=\frac{d}{dx}(e^x+1)[/Tex]

⇒ y’ = e^x

[Tex]\frac{d}{dx}(y’)=\frac{d}{dx}(e^x)[/Tex]

⇒ y” = e^x

Substituting the obtained values, we have

y” – y = e^x – e^x = 0 = RHS

Hence y is a solution of the given equation.

Question 2. y = x² + 2x + C : y’ – 2x – 2 = 0

Solution:

Given: y = x² + 2x + C, differentiating both sides w.r.t.x., we have:

[Tex]\frac{dy}{dx}=\frac{d}{dx}(x^2 + 2x + C)[/Tex]

⇒ y’ = 2x + 2

Substituting the obtained value, we have

y’ – 2x – 2 = (2x + 2) – 2x – 2 = 0 = RHS

Hence y is a solution of the given equation.

Question 3. y = cos x + C : y’ + sin x = 0

Solution:

Given: y = cos x + C, differentiating both sides w.r.t.x., we have:

[Tex]\frac{dy}{dx}=\frac{d}{dx}(cosx + C)[/Tex]

⇒ y’ = -sin x

Substituting the obtained value, we have

y’ + sin x = -sin x + sin x = 0 = RHS

Hence y is a solution of the given equation.

Question 4. y = [Tex]\sqrt{1+ x^2}[/Tex] : y’ = [Tex]\frac{xy}{1+x^2}[/Tex]

Solution:

Given: y = [Tex]\sqrt{1+x^2}[/Tex], differentiating both sides w.r.t.x., we have:

[Tex]\frac{dy}{dx}=\frac{d}{dx}(\sqrt{1+x^2})[/Tex]

= [Tex]\frac{1}{2\sqrt{1+x^2}}\times\frac{d}{dx}({1+x^2})[/Tex]

= [Tex]\frac{2x}{2\sqrt{1+x^2}}[/Tex]

= [Tex]\frac{x}{1+x^2}\times\sqrt{1+x^2}[/Tex]

= [Tex]\frac{x}{1+x^2}\times y[/Tex]

= [Tex]\frac{xy}{1+x^2}[/Tex]

= RHS

Hence y is a solution of the given equation.

Question 5. y = Ax : xy’ = y (x ≠ 0)

Solution:

Given: y = Ax, differentiating both sides w.r.t.x., we have:

[Tex]\frac{dy}{dx}=\frac{d}{dx}(Ax)[/Tex]

⇒ y’ = A

Substituting the obtained value, we have

xy’ = xA = Ax = y = RHS

Hence y is a solution of the given equation.

Question 6. y = x sin x : xy’ = y + x[Tex]\sqrt{x^2 − y^2}[/Tex] (x ≠ 0 and x > y) or x < – y

Solution:

Given: y = x sin x, differentiating both sides w.r.t.x., we have:

\frac{dy}{dx}=\frac{d}{dx}(x sin x)

⇒ y’ = [Tex]sin\space x.\frac{d}{dx}(x)+x.\frac{d}{dx}(sin\space x)[/Tex]

= sin x + x cos x

Substituting the obtained value, we have

xy’ = x(sin x + x cos x)

= x sin x + x² cos x

= y + x².[Tex]\sqrt{1-sin^2x}[/Tex]

= y + x².[Tex]\sqrt{1-(\frac{y}{x})^2}[/Tex]

= [Tex]y + x\sqrt{x^2-y^2}[/Tex]

= RHS

Hence y is a solution of the given equation.

Question 7. xy = log y + C : y’ = [Tex]\frac{y^2}{1-xy}[/Tex]

Solution:

Given: y = log y + C, differentiating both sides w.r.t.x., we have:

[Tex]\frac{d}{dx}(xy)=\frac{d}{dx}(log\space y)[/Tex]

⇒ [Tex]y’ = y\frac{d}{dx}(x)+x.\frac{dy}{dx}=\frac{1}{x}\frac{dy}{dx}[/Tex]

= [Tex]y + xy’ = \frac{1}{y}.y'[/Tex]

= y² + xyy’ = y’

= (xy – 1)y’ = -y²

y’ = [Tex]\frac{y^2}{1-xy}[/Tex]

LHS = RHS

Hence y is a solution of the given equation.

Question 8. y – cos y = x : (y sin y + cos y + x)y’ = y

Solution:

Given: y – cos y = x, differentiating both sides w.r.t.x., we have:

[Tex]\frac{dy}{dx}-\frac{d}{dx}(cos\space y)=\frac{d}{dx}(x)[/Tex]

⇒ y’ – (-sin y).y’ = 1

[Tex]⇒ y’ = \frac{1}{1+sin\space y}[/Tex]

Substituting the obtained value, we have

(y sin y + cos y + x)y’ = [Tex](y sin y + cos y + x) \times \frac{1}{1+sin\space y}\\=y(1+siny).\frac{1}{1+siny}[/Tex]

= y

= RHS

Hence y is a solution of the given equation.

Question 9. x + y = tan^–1y : y²y’ + y² + 1 = 0

Solution:

Given: x + y = tan^–1y, differentiating both sides w.r.t.x., we have:

[Tex]\frac{d}{dx}(x+y)= \frac{d}{dx}(tan^{-1} y)\\ ⇒ 1 + y’ = \frac{y’}{1+y^2}\\ ⇒ y'[\frac{-y^2}{1+y^2}]=1\\ ⇒ y’ = \frac{-(1+y^2)}{y^2}[/Tex]

Substituting the obtained value, we have

y²y’ + y² + 1 = [Tex]y^2\frac{-(1+y^2)}{y^2} + y^2 + 1[/Tex]

= -1 – y² + y² + 1

= 0

= RHS

Hence y is a solution of the given equation.

Question 10. [Tex]y = \sqrt{a^2 − x^2} \space : \space x + y\frac{dy}{dx} = 0[/Tex]

Solution:

Given: y = [Tex]\sqrt{a^2 − x^2}[/Tex], differentiating both sides w.r.t.x., we have:

[Tex]\frac{dy}{dx}= \frac{d}{dx}(\sqrt{a^2 − x^2})\\⇒\frac{dy}{dx}=\frac{1}{2\sqrt{a^2-x^2}}.\frac{d}{dx}(a^2-x^2)\\⇒\frac{dy}{dx}=\frac{1}{2\sqrt{a^2-x^2}}.(-2x)\\⇒\frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}[/Tex]

Substituting the obtained value, we have

[Tex]x + y\frac{dy}{dx} = x + (\sqrt{a^2 − x^2}).(\frac{-x}{\sqrt{a^2-x^2}})[/Tex]

= x + (-x)

= 0

= RHS

Hence y is a solution of the given equation.

Question 11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Solution:

Option D: We know that the number of constants in the general solution of an n-order differential equation is equal to its order. As a result, the general equation for a fourth order differential equation contains four constants.

Question 12. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Solution:

Option D: The number of arbitrary constants in the particular solution of a third-order differential equation is zero because all constants are determined by the given conditions.

Related Articles-

• Chapter 9 Differential Equations-Exercise -9.1
• Chapter 9 Differential Equations-Exercise -9.3