Referring to the AP Inter 2nd Year Maths Study Material Chapter 9 Differential Equations Exercise 9b Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Differential Equations Solutions Exercise 9b
I.
Question 1.
Verify that y = ex + 1 is a solution to y” – y’= 0
Solution:
Given function is y = ex + 1; Differentiating both sides w.r.t x, we have
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(e^x+1\right)\) ⇒ y’ = ex ………..(1); Again differentiating both sides w.r.t x, we have
\(\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^x\right)\) ⇒ y” = ex …………(2)
From (1) and (2), y” – y’= ex – ex = 0
Thus, the given function is the solution of corresponding differential equation. Hence verified.
Question 2.
Verify that y= x2 + 2x + C is a solution to y’ – 2x – 2 = 0
Solution:
Given function is y = x2 + 2x + c ⇒ y’ = \(\frac{d}{d x}\)(x2 + 2x + c) ⇒ y’ = 2x + 2
∴ y’ – 2x – 2 = (2x + 2) – 2x – 2 = 0
Thus, the given function is the solution of corresponding differential equation. Hence verified.
Question 3.
Verify that y = cosx + C is a solution to y’ + sin x = 0
Solution:
Given function is y = cos x + C ⇒ y’ = \(\frac{d}{d x}\)(cos x + C) ⇒ y’ = -sin x
∴ y’ – sinx = -sinx + sinx = 0
Thus, the given function is the solution of corresponding differential equation. Hence verified.
Question 4.
Verify that y = \(\sqrt{1+x^2}\) is a solution to y’ = \(\frac{x y}{1+x^2}\)
Solution:
Given function is y = \(\sqrt{1+x^2} \Rightarrow y^{\prime}=\frac{d}{d x}\left(\sqrt{1+x^2}\right)=\frac{1}{2 \sqrt{1+x^2}} \cdot \frac{d}{d x}\left(1+x^2\right)\)
= \(\frac{2 x}{2 \sqrt{1+x^2}}=\frac{x}{2 \sqrt{1+x^2}}=\frac{x\left(\sqrt{1+x^2}\right)}{\left(1+x^2\right)}=\frac{x y}{1+x^2}\)
Thus, the given function is the solution of corresponding differential equation.Hence verified.
Question 5.
Verify that y = Ax is a solution to xy’ = y(x ≠ 0)
Solution:
Given function is y = Ax ⇒ y’ = \(\frac{d}{d x}\)(Ax) = A ∴ xy’ = xA = Ax = y
Thus, the given function is the solution of corresponding differential equation. Hence verified.
Question 6.
Verify that y = x sin x is a solution to xy’ = y + x\(\sqrt{x^2-y^2}\)(x ≠ 0 and x > y or x < -y)
Solution:
Given that y = x sinx ⇒ y’ = \(\frac{d}{d x}\)(x sinx) = sin x\(\frac{d}{d x}\)(x) + x\(\frac{d}{d x}\)(sinx) = sin x + x cosx
∴ xy’ = x(sin x + x cox x) = x sin x + x2 cox x [∵ y = x sin x ⇒ \(\frac{\mathrm{y}}{\mathrm{x}}\) = sinx]
= y + x2. \(\sqrt{1-\sin ^2 x}\) = y + x2\(\sqrt{1-\left(\frac{y}{x}\right)^2}\) = y + x\(\sqrt{x^2-y^2}\).
Hence verified.
Question 7.
Verify that xy = logy + C is a solution to y’ = \(\frac{y^2}{1-x y}\)(xy ≠ 1)
Solution:
Given that xy = log y + C ⇒ \(\frac{d}{d x}\)(xy + C) = \(\frac{d}{d x}\)(logy) ⇒ y\(\frac{d}{d x}\)(x) + x.\(\frac{d y}{d x}=\frac{1}{y} \frac{d y}{d x}\)
⇒ y + xy’ = \(\frac{1}{y}\).y’ ⇒ y2 + xyy’ = y’ ⇒ (xy – 1)y’ = -y2 ⇒ y ‘ = \(\frac{y^2}{1-x y}\). Hence verified
Question 8.
Verify that y – cos y = x is a solution to (y siny + cos y + x)y’ = y
Solution:
Given that y – cos y = x ⇒ \(\frac{d y}{d x}-\frac{d}{d x}\)(cos y) = \(\frac{d}{d x}\)(x)
⇒ y’ – (-sin y).y’ = 1 ⇒ y'(1 + sin y) = 1 ⇒ y’ = \(\frac{1}{1+\sin y}\) [∵ x = y – cos y]
∴ (y sin y + cos y + x)y’ = (y sin y + cos y + y – cos y) × \(\frac{1}{1+\sin y}\) = y(1 + sin y).\(\frac{1}{1+\sin y}\) = y
Hence verified.
Question 9.
Verify that x + y = tan-1 y is a solution to y2y’ + y2 + 1 = 0
Solution:
Given that x + y = tan-1 y ⇒ \(\frac{d}{d x}\)(x + y) = \(\frac{d}{d x}\)(tan-1y)
⇒ \(1+y^{\prime}=\left[\frac{1}{1+y^2}\right] y^{\prime} \Rightarrow y^{\prime}\left[\frac{1}{1+y^2}-1\right]=1\)
⇒ \(y^{\prime}\left[\frac{1-\left(1+y^2\right)}{1+y^2}\right]=1 \Rightarrow y^{\prime}\left[\frac{-y^2}{1+y^2}\right]=1 \Rightarrow y^{\prime}=\frac{-\left(1+y^2\right)}{y^2}\)
∴ y2y’ + y2 + 1 = y2\(\left[\frac{-\left(1+y^2\right)}{y^2}\right]\) + y2 + 1 = -1 – y2 + y2 + 1 = 0. Hence verified.
Question 10.
Verify that y = \(\sqrt{a^2-x^2}\); x (-a, a) is a solution to x + y\(\frac{d y}{d x}\) = 0(y ≠ 0)
Solution:
Given that y = \(\sqrt{a^2-x^2} \Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^2-x^2}\right) \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^2-x^2}} \cdot \frac{d}{d x}\left(a^2-x^2\right)\)
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{a^2-x^2}}(-2 x) \Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a^2-x^2}}\)
∴ x + y\(\frac{d y}{d x}=x+\sqrt{a^2-x^2} \times \frac{-x}{\sqrt{a^2-x^2}}\) = x – x = 0. Hence verified