Question

consider the differential equation
y''+ay'+y=sin x for x for xεR. (**) 
then which one of the following is true?

• A. If a=0, then all the solutions of (**) are unbounded over R
• B. If a=1, then all the solutions of (**) are unbounded over (0, ∞).
• C. If a=1, then all the solutions of (**) tend to zero as x→∞
• D. If a=2, then all the solutions of (**) are bounded over (-∞, 0).

Answer 1

The Correct Option is A

To determine which of the given statements about the differential equation y′′+ay′+y=sin⁡xy′′+ay′+y=sinx are true, let's analyze each option based on the differential equation's behavior for different values of \(a\).

1. First, consider the statement: If \(a=0\), then all the solutions of \(y'' + ay' + y = \sin x\) are unbounded over \(\mathbb{R}\).

For \(a = 0\), the differential equation simplifies to:

\(y'' + y = \sin x\)

This is a second-order linear non-homogeneous differential equation. The complementary solution ycyc​ to the homogeneous equation y′′+y=0y′′+y=0 is:

\(y_c = C_1 \cos x + C_2 \sin x\), where C1C1​ and C2C2​ are arbitrary constants.

The particular solution ypyp​ can be guessed for a sine function on the right-hand side, which typically takes the form Acos⁡x+Bsin⁡xAcosx+Bsinx. Plugging this guess into the equation allows us to solve for AA and BB, but since the right-hand side is sin⁡xsinx, and it is also in the complementary solution, the particular form is slightly different, usually involving terms like xcos⁡xxcosx or xsin⁡xxsinx. This results in:

\(y_p = -\frac{1}{2} x \cos x\)

The general solution yy would then be:

\(y = y_c + y_p = C_1 \cos x + C_2 \sin x - \frac{1}{2} x \cos x\)

From the particular solution component −12xcos⁡x−21​xcosx, as x→∞x→∞ or x→−∞x→−∞, this term grows unboundedly, making the solution unbounded over RR. Hence, this statement is true.

1. Consider the statement: If \(a=1\), then all the solutions of \(y'' + ay' + y = \sin x\) are unbounded over \((0, \infty)\).

For \(a = 1\), the differential equation is:

\(y'' + y' + y = \sin x\)

The complementary solution ycyc​ for the homogeneous equation y′′+y′+y=0y′′+y′+y=0 involves solving the characteristic equation r2+r+1=0r2+r+1=0. The roots are complex, leading to a complementary solution of:

\(y_c = e^{-\frac{1}{2}x}(C_1 \cos (\frac{\sqrt{3}}{2}x) + C_2 \sin (\frac{\sqrt{3}}{2}x))\)

Since the complementary part is oscillatory but decays to zero as x→∞x→∞, the long-term behavior is determined by ypyp​, which is bounded. Hence, it does not become unbounded in (0,∞)(0,∞), making this statement false.

1. For the statement: If \(a=1\), then all the solutions of \(y'' + ay' + y = \sin x\) tend to zero as x→∞x→∞.

Given our analysis from the previous point, the complementary solution decays to zero, and the particular solution is bounded as x→∞x→∞. Therefore, the statement holds true as well.

1. Consider the statement: If \(a=2\), then all the solutions of \(y'' + ay' + y = \sin x\) are bounded over (−∞,0)(−∞,0).

With \(a = 2\), we have:

\(y'' + 2y' + y = \sin x\)

The roots of the characteristic equation r2+2r+1=0r2+2r+1=0 are real and repeated. Thus, the complementary solution is:

\(y_c = (C_1 + C_2 x) e^{-x}\)

The complementary solution tends to zero as x→−∞x→−∞, and the particular solution for sin⁡xsinx on the right-hand side also remains bounded. Therefore, this statement is true.

In conclusion, the correct answer among the options is: If \(a=0\), then all the solutions of \(y'' + ay' + y = \sin x\) are unbounded over RR.