Ordinary Differential Equation T 3

Rsquared Mathematics

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Time: 45:00

Topic: Ordinary Differential Equation

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The Quiz contains a total of 10 Questions.
Part B(Single Correct) contains 5 questions, each of 3 marks and -0.75 marks for incorrect answers.
Part C(MSQ) contains 5 questions, each with 4.75 marks and no negative marking. The maximum score for the Test is 38.75. Best of Luck.

Part - B.

1. The solution of the differential equation

\(\frac{dx}{dt} - x + \frac{dy}{dt} + y = 0\)

\(2\frac{dx}{dt} + 2x + 2\frac{dy}{dt} - 2y = t\)

is

(a.) \(x(t) = \frac{t^2}{16} + \frac{t}{8} + c_1\)
\(y(t) = \frac{t^2}{16} - \frac{t}{8} + c_2\)
(b.) \(x(t) = \frac{t^2}{16} - \frac{t}{8} + c_1\)
\(y(t) = \frac{t^2}{16} + \frac{t}{8} + c_2\)
(c.) \(x(t) = \frac{t^2}{8} + \frac{t}{16} + c_1\)
\(y(t) = \frac{t^2}{8} - \frac{t}{16} + c_2\)
(d.) \(x(t) = \frac{t^2}{8} - \frac{t}{16} + c_1\)
\(y(t) = \frac{t^2}{8} + \frac{t}{16} + c_2\)

Explanation: Equation (1) is \((D-1) x + (D+1) y = 0\)

Equation (2) is \(2(D+1) x + 2(D-1) y = t\)

Equation (1) \(\times 2(D+1)\) - Equation (2) \(\times (D-1)\)

\[2(D+1)^2 - 2(D-1)^2 y = -(D-1) t\]

\[\Rightarrow 8Dy = (1-D)t\]

\[\Rightarrow y = \frac{1}{8D}(1-D)t = \frac{1}{8}\cdot\frac{t^2}{2} - \frac{t}{8} + c_1\]

\[\Rightarrow y(t) = \frac{t^2}{16} - \frac{t}{8} + c_1\]

2. Given that \(y_1(x) = x\) is a solution of the differentiable equation \(x^2 y^{\prime\prime} + 2xy^\prime - 2y = 0\).

Then a linearly independent solution \(y_2(x)\) is given by -

(a.) \(x^2\)
(b.) \(\frac{1}{x^2}\)
(c.) \(\frac{1}{x^3}\)
(d.) \(x^4\)

Explanation: Let \(y_2(x) = x \cdot v(x)\) be a solution of \(x^2 y^{\prime\prime} + 2xy^\prime - 2y = 0\)

\(y_2^\prime = x \cdot v^\prime + v \quad \Rightarrow \quad y_2^{\prime\prime} = x \cdot v^{\prime\prime} + 2v^\prime\)

\(\Rightarrow 0 = x^3 v^{\prime\prime}(x) + 4x^2 v^\prime(x)\)

\(\Rightarrow x \cdot v^{\prime\prime}(x) + \left(2 + \frac{2}{x}x\right) v^\prime(x) = 0\)

\(\Rightarrow \frac{w^\prime(x)}{w(x)} = \frac{-4}{x} \quad \left[\text{If } w(x) = v^\prime(x)\right]\)

\(\Rightarrow w(x) = c_1x^{-4}\)

Therefore, \(v(x) = c_2\left(\frac{1}{x^3}\right) + c_3\)

\(\Rightarrow y_2(x) = c_2\left(\frac{1}{x^2}\right) + c_3x\)

3. The particular solution of

\[\left(D^2-4 D+4\right) y = 8x^2 e^{2x} \sin(2x)\]

is

(a.) Periodic
(b.) Strictly increasing
(c.) Strictly decreasing
(d.) Not monotonic

Explanation:

\(y_p(x) = P \cdot I = e^{2x}\left[\left(3-2x^2\right)\sin(2x) - 4x\cos(2x)\right]\)

4. Consider the differential equation \(\sin(x) \frac{d^2 y}{d x^2} - \frac{d y}{d x} + \sin(x) y = 0, x \in \left(\frac{\pi}{6}, \frac{\pi}{3}\right)\)

Let \(v(x)\) and \(u(x)\) be its linearly independent solutions. Then which of the following is not true about \(W(x)\)?

(a.) \(W(x)\) is increasing
(b.) \(W(x)\) is Riemann integrable
(c.) \(W(x)\) holds IVP
(d.) \(W(x)\) is one-one

Explanation: Find \(W(x)\) and observe.

5. Consider the differential equation \(\frac{dx}{dt} = \frac{x^3 - x}{1 + e^x}\) with \(x(0) = 0\)

Then which of the following is correct -

(a.) \(\lim_{t \to \infty} x(t) = 0\)
(b.) \(\lim_{t \to \infty} x(t) = \frac{1}{2}\)
(c.) \(\lim_{t \to \infty} x(t) = \frac{1}{3}\)
(d.) \(\lim_{t \to \infty} x(t) = 1\)

Explanation: It has only the trivial solution.

Part - C

6. The particular solution of \(y^{\prime \prime}+y=\sin(2x)\sin(x)\) satisfies

(a.) \(y_p\left(\frac{\pi}{2}\right)=\frac{\pi}{8}\)
(b.) \(y_p(0)=\frac{1}{16}\)
(c.) \(y_p^{\prime}(0)=0\)
(d.) \(y_p^{\prime}\left(\frac{\pi}{2}\right)=\frac{7}{16}\)

Explanation: \(y_p(x)=\frac{x}{4} \sin(x)+\frac{1}{16} \cos(3x)\)

7. Consider the IVP \(y^{\prime}=\frac{10}{3}xy^{2/5}; y(0)=-1\)

(a.) \(\exists\) a unique solution \(y(x)\) to the IVP on some open interval that contains \(x_0=0\)
(b.) \(\exists\) infinitely many solutions \(y(x)\) to the IVP on some open interval that contains \(x_0=0\)
(c.) The largest open interval on which \(y(x)\) is unique is \((-1,1)\)
(d.) The largest open interval on which \(y(x)\) is unique is \(\left(\frac{-1}{2}, \frac{1}{2}\right)\)

Explanation: \(y^{\prime}=\frac{10}{3}xy^{2/5} \Rightarrow y(x)=(x^2+c)^{5/3}\)
\(y(0)=-1 \Rightarrow c=-1\)
\(\therefore y(x)=(x^2-1)^{5/3}\)
\(\& \because y(x)\) is continuous, \(\exists\) an open interval \(I\) containing \(x_0=0\) on which \(y\) has no zeroes \(\&\) hence \(I=(-1,1)\) on which \(y(x)\) is a unique solution.

8. Let \(y(x)\) be the solution of the ODE \(y^{\prime \prime}-y=-1\) which passes through the origin and remains bounded as \(x \rightarrow \infty\). Then, which of the following statements is/are true?

(a.) \(y(-1)=1-e\)
(b.) \(y(-1)=0\)
(c.) \(\lim _{x \rightarrow 0} \frac{y(x)}{x}=0\)
(d.) \(\lim _{x \rightarrow 0} \frac{y(x)}{x}=1\)

Explanation: Solution is \(y(x)=c_1 e^x+c_2 e^{-x}+1\)

\(\Rightarrow c_1=0, c_2=1\)

\(\Rightarrow y(x)=1-e^{-x}\)

\(y(-1)=1-e, \lim _{x \rightarrow 0} \frac{y(x)}{x}=1\)

9. Let \(W\) denote the Wronskian of two solutions of \(z^{\prime}=\begin{bmatrix}1 & 1 \\ 4 & 1\end{bmatrix} z\). Then \(W\) can be

(a.) Zero
(b.) \(-4 e^{2 t}\)
(c.) \(4 e^{2 t}\)
(d.) \(-2 e^{2 t}\)

Explanation: If the solutions are linearly dependent, then \(W=\mathbf{0}\). Otherwise, the two linearly independent solutions are:

\(\begin{bmatrix}e^{3 t} \\ 2 e^{3 t}\end{bmatrix}\) and \(\begin{bmatrix}e^{-t} \\ -2 e^{-t}\end{bmatrix}\)

Then \(W=\begin{vmatrix}e^{3 t} & e^{-t} \\ 2 e^{3 t} & -2 e^{-t}\end{vmatrix}=-2 e^{2 t}-2 e^{2 t}=-4 e^{2 t}\)

10. The initial value problem \(x^2 y^{\prime}=1-x^2+y^2-x^2 y^2\) with \(y(1)=\alpha\) has

(a.) No solution if \(\alpha=0\)
(b.) Unique solution if \(\alpha=0\)
(c.) Infinite number of solutions if \(\alpha=0\)
(d.) Unique solution if \(\alpha=1\)

Explanation:

\(x^2 y^{\prime}=(1-x^2)(1+y^2) \Rightarrow \frac{dy}{1+y^2}=\left(\frac{1}{x^2}-1\right)dx\)

\(\Rightarrow y(x)=\tan\left(-\frac{1}{x}-x+c\right)\)