Higher-Order ODEs: Practice Questions & Solutions

Higher-Order Ordinary Differential Equations — Practice Questions & Solutions

> Perfect for: Semester exam preparation, competitive exams (GATE, ESE), and concept reinforcement.

Module 1: Linear Differential Equation of nth Order

Q1. Solve: $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$

Solution:

This is a second-order linear homogeneous ODE with constant coefficients.

Step 1: Write the auxiliary equation by substituting $D = \frac{d}{dx}$:

$(D^2 - 5D + 6)y = 0$

$m^2 - 5m + 6 = 0$

Step 2: Solve the quadratic:

$(m - 2)(m - 3) = 0$

$m_1 = 2, \quad m_2 = 3$

Step 3: Since roots are real and distinct, the complementary function is:

$\boxed{y_c = C_1 e^{2x} + C_2 e^{3x}}$

Since $f(x) = 0$, the particular integral $y_p = 0$.

Final Answer: $y = C_1 e^{2x} + C_2 e^{3x}$

Q2. Solve: $\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 4y = 0$

Solution:

Step 1: Auxiliary equation:

$m^2 + 4m + 4 = 0$

$(m + 2)^2 = 0$

$m = -2, -2 \quad \text{(repeated root)}$

Step 2: For real and equal roots, the solution is:

$\boxed{y = (C_1 + C_2 x)e^{-2x}}$

Q3. Solve: $\frac{d^2y}{dx^2} + 4y = 0$

Solution:

Step 1: Auxiliary equation:

$m^2 + 4 = 0$

$m^2 = -4$

$m = \pm 2i \quad \text{(complex conjugate roots, } \alpha = 0, \beta = 2\text{)}$

Step 2: For complex roots $\alpha \pm i\beta$:

$\boxed{y = e^{0x}(C_1 \cos 2x + C_2 \sin 2x) = C_1 \cos 2x + C_2 \sin 2x}$

Q4. Solve: $\frac{d^3y}{dx^3} - 6\frac{d^2y}{dx^2} + 11\frac{dy}{dx} - 6y = 0$

Solution:

Step 1: Auxiliary equation:

$m^3 - 6m^2 + 11m - 6 = 0$

Step 2: By trial, $m = 1$ is a root. Factorize:

$(m - 1)(m^2 - 5m + 6) = 0$

$(m - 1)(m - 2)(m - 3) = 0$

$m = 1, 2, 3$

Step 3: All roots real and distinct:

$\boxed{y = C_1 e^x + C_2 e^{2x} + C_3 e^{3x}}$

Q5. Solve: $\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 2y = e^{3x}$

Solution:

Step 1: Auxiliary equation:

$m^2 - 3m + 2 = 0$

$(m - 1)(m - 2) = 0$

$m = 1, 2$

Complementary Function:

$y_c = C_1 e^x + C_2 e^{2x}$

Step 2: Find Particular Integral using operator method:

$y_p = \frac{1}{D^2 - 3D + 2} e^{3x}$

Substitute $D = 3$:

$y_p = \frac{1}{9 - 9 + 2} e^{3x} = \frac{1}{2}e^{3x}$

Final Answer:

$\boxed{y = C_1 e^x + C_2 e^{2x} + \frac{1}{2}e^{3x}}$

Q6. Solve: $\frac{d^2y}{dx^2} + y = \sin x$

Solution:

Step 1: Auxiliary equation:

$m^2 + 1 = 0 \Rightarrow m = \pm i$

$y_c = C_1 \cos x + C_2 \sin x$

Step 2: Particular Integral (resonance case — $f(x) = \sin x$ matches CF):

$y_p = \frac{1}{D^2 + 1} \sin x$

Since $D^2 + 1 = 0$ when $D^2 = -1$, we use the rule for failure case:

$y_p = x \cdot \frac{1}{2D} \sin x = \frac{x}{2} \int \sin x \, dx = -\frac{x}{2}\cos x$

Final Answer:

$\boxed{y = C_1 \cos x + C_2 \sin x - \frac{x}{2}\cos x}$

Q7. Solve: $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = x^2 e^x$

Solution:

Step 1: Auxiliary equation:

$m^2 - 2m + 1 = 0 \Rightarrow (m-1)^2 = 0 \Rightarrow m = 1, 1$

$y_c = (C_1 + C_2 x)e^x$

Step 2: Particular Integral:

$y_p = \frac{1}{(D-1)^2} x^2 e^x = e^x \frac{1}{D^2} x^2 = e^x \cdot \frac{x^4}{12}$

Final Answer:

$\boxed{y = (C_1 + C_2 x)e^x + \frac{x^4}{12}e^x}$

Q8. Solve: $\frac{d^3y}{dx^3} - 7\frac{dy}{dx} + 6y = 1 + x + x^2$

Solution:

Step 1: Auxiliary equation:

$m^3 - 7m + 6 = 0$

By trial $m = 1$: $1 - 7 + 6 = 0$ ✓

$(m - 1)(m^2 + m - 6) = 0$

$(m - 1)(m + 3)(m - 2) = 0$

$m = 1, 2, -3$

$y_c = C_1 e^x + C_2 e^{2x} + C_3 e^{-3x}$

Step 2: Particular Integral for polynomial:

$y_p = \frac{1}{D^3 - 7D + 6}(1 + x + x^2)$

Using binomial expansion or long division:

$y_p = \frac{1}{6}\left(1 + \frac{7D - D^3}{6}\right)^{-1}(1 + x + x^2)$

$= \frac{1}{6}\left(1 - \frac{7D}{6} + \frac{49D^2}{36} + \cdots\right)(1 + x + x^2)$

Apply to $1 + x + x^2$:

• $D(1 + x + x^2) = 1 + 2x$
• $D^2(1 + x + x^2) = 2$

$y_p = \frac{1}{6}\left[(1 + x + x^2) - \frac{7}{6}(1 + 2x) + \frac{49}{36}(2)\right]$

$= \frac{1}{6}\left[1 + x + x^2 - \frac{7}{6} - \frac{14x}{6} + \frac{98}{36}\right]$

Simplify:

$y_p = \frac{x^2}{6} - \frac{4x}{36} + \frac{50}{216} = \frac{x^2}{6} - \frac{x}{9} + \frac{25}{108}$

Final Answer:

$\boxed{y = C_1 e^x + C_2 e^{2x} + C_3 e^{-3x} + \frac{x^2}{6} - \frac{x}{9} + \frac{25}{108}}$

Module 2: Simultaneous Linear Differential Equations

Q9. Solve the simultaneous equations:

$\frac{dx}{dt} + y = \sin t$

$\frac{dy}{dt} + x = \cos t$

Solution:

Step 1: Write in operator form ($D = \frac{d}{dt}$):

$(1) \quad Dx + y = \sin t$ $(2) \quad x + Dy = \cos t$

Step 2: Eliminate $y$. From (1): $y = \sin t - Dx$

Substitute into (2):

$x + D(\sin t - Dx) = \cos t$

$x + \cos t - D^2x = \cos t$

$(1 - D^2)x = 0$

$D^2x - x = 0 \Rightarrow (D^2 - 1)x = 0$

Step 3: Solve for $x$:

$m^2 - 1 = 0 \Rightarrow m = \pm 1$

$x = C_1 e^t + C_2 e^{-t}$

Step 4: Find $y$ from equation (1):

$y = \sin t - Dx = \sin t - (C_1 e^t - C_2 e^{-t})$

$y = \sin t - C_1 e^t + C_2 e^{-t}$

Final Answer:

$\boxed{x = C_1 e^t + C_2 e^{-t}, \quad y = \sin t - C_1 e^t + C_2 e^{-t}}$

Q10. Solve:

$\frac{dx}{dt} + 2x - 3y = t$

$\frac{dy}{dt} - 3x + 2y = e^{2t}$

Solution:

Step 1: Operator form:

$(D + 2)x - 3y = t \quad \text{...(1)}$

$-3x + (D + 2)y = e^{2t} \quad \text{...(2)}$

Step 2: Eliminate $y$. Multiply (1) by $(D+2)$ and (2) by $3$:

$(D+2)^2 x - 3(D+2)y = (D+2)t = 1 + 2t$

$-9x + 3(D+2)y = 3e^{2t}$

Add:

$(D+2)^2 x - 9x = 1 + 2t + 3e^{2t}$

$(D^2 + 4D + 4 - 9)x = 1 + 2t + 3e^{2t}$

$(D^2 + 4D - 5)x = 1 + 2t + 3e^{2t}$

Step 3: Solve homogeneous part:

$m^2 + 4m - 5 = 0 \Rightarrow (m+5)(m-1) = 0 \Rightarrow m = 1, -5$

$x_c = C_1 e^t + C_2 e^{-5t}$

Step 4: Particular Integral:

$x_p = \frac{1}{D^2 + 4D - 5}(1 + 2t + 3e^{2t})$

For $1 + 2t$:

$\frac{1}{-5}\left(1 - \frac{4D + D^2}{5}\right)^{-1}(1 + 2t) = -\frac{1}{5}\left(1 + \frac{4D}{5}\right)(1 + 2t)$

$= -\frac{1}{5}\left(1 + 2t + \frac{8}{5}\right) = -\frac{1}{5}\left(2t + \frac{13}{5}\right) = -\frac{2t}{5} - \frac{13}{25}$

For $3e^{2t}$:

$\frac{3e^{2t}}{4 + 8 - 5} = \frac{3e^{2t}}{7}$

$x_p = -\frac{2t}{5} - \frac{13}{25} + \frac{3e^{2t}}{7}$

Step 5: Find $y$ from equation (1):

$3y = (D+2)x - t$

Compute $(D+2)x$:

$x = C_1 e^t + C_2 e^{-5t} - \frac{2t}{5} - \frac{13}{25} + \frac{3e^{2t}}{7}$

$Dx = C_1 e^t - 5C_2 e^{-5t} - \frac{2}{5} + \frac{6e^{2t}}{7}$

$(D+2)x = 3C_1 e^t - 3C_2 e^{-5t} - \frac{2}{5} - \frac{4t}{5} - \frac{26}{25} + \frac{9e^{2t}}{7}$

$3y = 3C_1 e^t - 3C_2 e^{-5t} - \frac{4t}{5} - \frac{36}{25} + \frac{9e^{2t}}{7} - t$

$\boxed{y = C_1 e^t - C_2 e^{-5t} - \frac{3t}{5} - \frac{12}{25} + \frac{3e^{2t}}{7}}$

Module 3: Second-Order Linear ODEs with Variable Coefficients

Q11. Solve: $x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} - 4y = 0$

Solution:

This is a Cauchy-Euler equation. Substitute $x = e^t$ (or $t = \ln x$):

$x\frac{dy}{dx} = Dy, \quad x^2\frac{d^2y}{dx^2} = D(D-1)y$

where $D = \frac{d}{dt}$.

Step 1: Transform:

$[D(D-1) + D - 4]y = 0$

$(D^2 - D + D - 4)y = 0$

$(D^2 - 4)y = 0$

Step 2: Solve:

$m^2 - 4 = 0 \Rightarrow m = \pm 2$

$y = C_1 e^{2t} + C_2 e^{-2t}$

Step 3: Substitute back $t = \ln x$, so $e^t = x$:

$\boxed{y = C_1 x^2 + \frac{C_2}{x^2}}$

Q12. Solve: $x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} + y = 0$

Solution:

Step 1: Substitute $x = e^t$:

$[D(D-1) + 3D + 1]y = 0$

$(D^2 - D + 3D + 1)y = 0$

$(D^2 + 2D + 1)y = 0$

$(D+1)^2 y = 0$

Step 2: Repeated root $m = -1, -1$:

$y = (C_1 + C_2 t)e^{-t}$

Step 3: Substitute $t = \ln x$:

$\boxed{y = \frac{C_1 + C_2 \ln x}{x}}$

Module 4: Solution by Changing the Independent Variable

Q13. Solve: $\frac{d^2y}{dx^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0$

Solution:

Step 1: Compare with standard form $\frac{d^2y}{dx^2} + P\frac{dy}{dx} + Qy = R$:

$P = \tan x, \quad Q = \cos^2 x, \quad R = 0$

Step 2: Choose new independent variable $z$ such that:

$\left(\frac{dz}{dx}\right)^2 = Q = \cos^2 x$

$\frac{dz}{dx} = \cos x \Rightarrow z = \sin x$

Step 3: Transform the equation. We have:

$\frac{d^2y}{dz^2} + P_1 \frac{dy}{dz} + Q_1 y = R_1$

where:

$Q_1 = \frac{Q}{(dz/dx)^2} = \frac{\cos^2 x}{\cos^2 x} = 1$

$P_1 = \frac{\frac{d^2z}{dx^2} + P\frac{dz}{dx}}{(dz/dx)^2} = \frac{-\sin x + \tan x \cdot \cos x}{\cos^2 x} = \frac{-\sin x + \sin x}{\cos^2 x} = 0$

Step 4: The transformed equation is:

$\frac{d^2y}{dz^2} + y = 0$

Step 5: Solve:

$m^2 + 1 = 0 \Rightarrow m = \pm i$

$y = C_1 \cos z + C_2 \sin z$

Final Answer:

$\boxed{y = C_1 \cos(\sin x) + C_2 \sin(\sin x)}$

Module 5: Method of Variation of Parameters

Q14. Solve by variation of parameters: $\frac{d^2y}{dx^2} + y = \tan x$

Solution:

Step 1: Complementary function:

$m^2 + 1 = 0 \Rightarrow m = \pm i$

$y_c = C_1 \cos x + C_2 \sin x$

So $y_1 = \cos x$, $y_2 = \sin x$.

Step 2: Compute Wronskian:

$W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1$

Step 3: Find particular solution:

$y_p = -y_1 \int \frac{y_2 f(x)}{W} dx + y_2 \int \frac{y_1 f(x)}{W} dx$

where $f(x) = \tan x$.

First integral:

$\int \frac{\sin x \cdot \tan x}{1} dx = \int \frac{\sin^2 x}{\cos x} dx = \int \frac{1 - \cos^2 x}{\cos x} dx$

$= \int (\sec x - \cos x) dx = \ln|\sec x + \tan x| - \sin x$

Second integral:

$\int \frac{\cos x \cdot \tan x}{1} dx = \int \sin x \, dx = -\cos x$

Step 4: Compute $y_p$:

$y_p = -\cos x [\ln|\sec x + \tan x| - \sin x] + \sin x [-\cos x]$

$= -\cos x \ln|\sec x + \tan x| + \cos x \sin x - \sin x \cos x$

$= -\cos x \ln|\sec x + \tan x|$

Final Answer:

$\boxed{y = C_1 \cos x + C_2 \sin x - \cos x \ln|\sec x + \tan x|}$

Q15. Solve by variation of parameters: $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = \frac{e^x}{x}$

Solution:

Step 1: CF: $m^2 - 2m + 1 = 0 \Rightarrow (m-1)^2 = 0 \Rightarrow m = 1, 1$

For repeated roots in variation of parameters, we use:

$y_1 = e^x, \quad y_2 = xe^x$

Step 2: Wronskian:

$W = \begin{vmatrix} e^x & xe^x \\ e^x & e^x + xe^x \end{vmatrix} = e^x(e^x + xe^x) - xe^x \cdot e^x = e^{2x}$

Step 3: Particular integral:

$y_p = -e^x \int \frac{xe^x \cdot \frac{e^x}{x}}{e^{2x}} dx + xe^x \int \frac{e^x \cdot \frac{e^x}{x}}{e^{2x}} dx$

$= -e^x \int 1 \, dx + xe^x \int \frac{1}{x} dx$

$= -xe^x + xe^x \ln|x|$

Final Answer:

$\boxed{y = (C_1 + C_2 x)e^x - xe^x + xe^x \ln|x| = (C_1 + C_2 x)e^x + xe^x(\ln|x| - 1)}$

Module 6: Cauchy-Euler Equation

Q16. Solve: $x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = \sin(\ln x)$

Solution:

Step 1: Substitute $x = e^t$, so $t = \ln x$:

$[D(D-1) + D + 1]y = \sin t$

$(D^2 + 1)y = \sin t$

Step 2: CF: $m^2 + 1 = 0 \Rightarrow m = \pm i$

$y_c = C_1 \cos t + C_2 \sin t$

Step 3: PI (resonance case):

$y_p = \frac{1}{D^2 + 1} \sin t = -\frac{t}{2}\cos t$

Step 4: Substitute $t = \ln x$:

$\boxed{y = C_1 \cos(\ln x) + C_2 \sin(\ln x) - \frac{\ln x}{2}\cos(\ln x)}$

Q17. Solve: $x^2\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} - 4y = x^2$

Solution:

Step 1: Substitute $x = e^t$:

$[D(D-1) - 2D - 4]y = e^{2t}$

$(D^2 - 3D - 4)y = e^{2t}$

Step 2: CF: $m^2 - 3m - 4 = 0 \Rightarrow (m-4)(m+1) = 0$

$y_c = C_1 e^{4t} + C_2 e^{-t}$

Step 3: PI:

$y_p = \frac{1}{D^2 - 3D - 4} e^{2t} = \frac{e^{2t}}{4 - 6 - 4} = -\frac{e^{2t}}{6}$

Step 4: Substitute back:

$\boxed{y = C_1 x^4 + \frac{C_2}{x} - \frac{x^2}{6}}$

Q18. Solve: $(x^2D^2 + xD + 1)y = \ln x \cdot \sin(\ln x)$

Solution:

Step 1: Substitute $x = e^t$:

$(D^2 + 1)y = t \sin t$

Step 2: CF: $y_c = C_1 \cos t + C_2 \sin t$

Step 3: PI using complex method:

Consider $\frac{1}{D^2 + 1} t e^{it}$ and take imaginary part.

Using $\frac{1}{D^2+1} = \frac{1}{(D+i)(D-i)}$ and shifting:

$\frac{1}{D^2+1} t e^{it} = e^{it} \frac{1}{(D+i)^2 + 1} t = e^{it} \frac{1}{D^2 + 2iD} t$

$= e^{it} \frac{1}{2iD}\left(1 + \frac{D}{2i}\right)^{-1} t = e^{it} \frac{1}{2iD}\left(t - \frac{1}{2i}\right)$

$= e^{it} \frac{1}{2i}\left(\frac{t^2}{2} - \frac{t}{2i}\right) = e^{it}\left(-\frac{it^2}{4} + \frac{t}{4}\right)$

$= (\cos t + i\sin t)\left(\frac{t}{4} - \frac{it^2}{4}\right)$

Imaginary part:

$= -\frac{t^2}{4}\cos t + \frac{t}{4}\sin t$

Step 4: Substitute $t = \ln x$:

$\boxed{y = C_1 \cos(\ln x) + C_2 \sin(\ln x) - \frac{(\ln x)^2}{4}\cos(\ln x) + \frac{\ln x}{4}\sin(\ln x)}$

Module 7: Applications of Differential Equations

Q19. Mechanical Vibrations — Free Damped Motion

A mass of $2$ kg is attached to a spring with constant $k = 128$ N/m. The damping coefficient is $c = 16$ N·s/m. The mass is displaced $0.1$ m and released. Find the equation of motion.

Solution:

Step 1: The equation of motion is:

$m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0$

$2\frac{d^2x}{dt^2} + 16\frac{dx}{dt} + 128x = 0$

$\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 64x = 0$

Step 2: Auxiliary equation:

$m^2 + 8m + 64 = 0$

$m = \frac{-8 \pm \sqrt{64 - 256}}{2} = \frac{-8 \pm \sqrt{-192}}{2} = -4 \pm 4\sqrt{3}i$

Step 3: Underdamped solution:

$x(t) = e^{-4t}(C_1 \cos 4\sqrt{3}t + C_2 \sin 4\sqrt{3}t)$

Step 4: Apply initial conditions $x(0) = 0.1$, $x'(0) = 0$:

At $t = 0$: $0.1 = C_1$

$x'(t) = -4e^{-4t}(C_1 \cos 4\sqrt{3}t + C_2 \sin 4\sqrt{3}t) + e^{-4t}(-4\sqrt{3}C_1 \sin 4\sqrt{3}t + 4\sqrt{3}C_2 \cos 4\sqrt{3}t)$

At $t = 0$: $0 = -4C_1 + 4\sqrt{3}C_2$

$C_2 = \frac{C_1}{\sqrt{3}} = \frac{0.1}{\sqrt{3}} = \frac{\sqrt{3}}{30}$

Final Answer:

$\boxed{x(t) = e^{-4t}\left(0.1 \cos 4\sqrt{3}t + \frac{\sqrt{3}}{30} \sin 4\sqrt{3}t\right) \text{ m}}$

Q20. LCR Circuit

An LCR circuit has $L = 0.5$ H, $R = 6$ Ω, $C = 0.02$ F, and $E(t) = 24\sin 10t$ V. Find the charge given , .

Solution:

Step 1: Circuit equation:

$L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = E(t)$

$0.5\frac{d^2q}{dt^2} + 6\frac{dq}{dt} + 50q = 24\sin 10t$

$\frac{d^2q}{dt^2} + 12\frac{dq}{dt} + 100q = 48\sin 10t$

Step 2: CF:

$m^2 + 12m + 100 = 0$

$m = \frac{-12 \pm \sqrt{144 - 400}}{2} = -6 \pm 8i$

$q_c = e^{-6t}(C_1 \cos 8t + C_2 \sin 8t)$

Step 3: PI:

$q_p = \frac{1}{D^2 + 12D + 100} 48\sin 10t$

Substitute $D^2 = -100$:

$q_p = \frac{48\sin 10t}{12D} = \frac{4\sin 10t}{D} = 4 \int \sin 10t \, dt = -\frac{4\cos 10t}{10} = -\frac{2\cos 10t}{5}$

Wait — check: $D^2 + 100 = 0$ at $D^2 = -100$, so we need to use the derivative rule:

$q_p = \frac{48\sin 10t}{12D} = \frac{4}{D}\sin 10t = -\frac{4\cos 10t}{10} = -0.4\cos 10t$

Actually, let's verify: $\frac{1}{D^2 + 12D + 100}\sin 10t$ with $D^2 = -100$:

$= \frac{1}{12D}\sin 10t = \frac{1}{12} \cdot \left(-\frac{\cos 10t}{10}\right) = -\frac{\cos 10t}{120}$

So $q_p = 48 \times \left(-\frac{\cos 10t}{120}\right) = -\frac{48\cos 10t}{120} = -\frac{2\cos 10t}{5} = -0.4\cos 10t$

Step 4: Apply ICs $q(0) = 0$, $q'(0) = 0$:

$q(t) = e^{-6t}(C_1 \cos 8t + C_2 \sin 8t) - 0.4\cos 10t$

At $t = 0$: $0 = C_1 - 0.4 \Rightarrow C_1 = 0.4$

$q'(t) = -6e^{-6t}(C_1 \cos 8t + C_2 \sin 8t) + e^{-6t}(-8C_1 \sin 8t + 8C_2 \cos 8t) + 4\sin 10t$

At $t = 0$: $0 = -6C_1 + 8C_2 \Rightarrow C_2 = \frac{6C_1}{8} = \frac{6 \times 0.4}{8} = 0.3$

Final Answer:

$\boxed{q(t) = e^{-6t}(0.4\cos 8t + 0.3\sin 8t) - 0.4\cos 10t}$

Q21. Simple Harmonic Motion — Spring-Mass System

A $4$ kg mass stretches a spring by $0.392$ m. The mass is set in motion from equilibrium with an upward velocity of $0.4$ m/s. Find the position function.

Solution:

Step 1: Find spring constant $k$:

$mg = k \cdot \Delta L \Rightarrow 4 \times 9.8 = k \times 0.392$

$k = \frac{39.2}{0.392} = 100 \text{ N/m}$

Step 2: Equation of motion (no damping, no external force):

$4\frac{d^2x}{dt^2} + 100x = 0$

$\frac{d^2x}{dt^2} + 25x = 0$

Step 3: Solve:

$m^2 + 25 = 0 \Rightarrow m = \pm 5i$

$x(t) = C_1 \cos 5t + C_2 \sin 5t$

Step 4: Apply ICs: $x(0) = 0$ (equilibrium), $x'(0) = -0.4$ (upward = negative):

$0 = C_1$

$x'(t) = -5C_1 \sin 5t + 5C_2 \cos 5t$

$-0.4 = 5C_2 \Rightarrow C_2 = -0.08$

Final Answer:

$\boxed{x(t) = -0.08\sin 5t \text{ m}}$

Q22. Forced Vibrations — Resonance

Solve: $\frac{d^2x}{dt^2} + 9x = 6\cos 3t$ with , .

Solution:

Step 1: CF: $m^2 + 9 = 0 \Rightarrow m = \pm 3i$

$x_c = C_1 \cos 3t + C_2 \sin 3t$

Step 2: PI (resonance — driving frequency = natural frequency):

$x_p = \frac{1}{D^2 + 9} 6\cos 3t = 6 \cdot \frac{t}{2 \cdot 3}\sin 3t = t\sin 3t$

(Using $\frac{1}{D^2+a^2}\cos at = \frac{t}{2a}\sin at$)

Step 3: General solution:

$x(t) = C_1 \cos 3t + C_2 \sin 3t + t\sin 3t$

Step 4: Apply ICs:

$x(0) = C_1 = 0$

$x'(t) = -3C_1 \sin 3t + 3C_2 \cos 3t + \sin 3t + 3t\cos 3t$

$x'(0) = 3C_2 = 0 \Rightarrow C_2 = 0$

Final Answer:

$\boxed{x(t) = t\sin 3t}$

> Note: The amplitude grows without bound — this is pure resonance.

Q23. Beats Phenomenon

Solve: $\frac{d^2x}{dt^2} + 25x = 4\cos 4.5t$ with , .

Solution:

Step 1: CF: $m^2 + 25 = 0 \Rightarrow m = \pm 5i$

$x_c = C_1 \cos 5t + C_2 \sin 5t$

Step 2: PI:

$x_p = \frac{1}{D^2 + 25} 4\cos 4.5t = \frac{4\cos 4.5t}{-20.25 + 25} = \frac{4\cos 4.5t}{4.75} = \frac{16\cos 4.5t}{19}$

Step 3: General solution:

$x(t) = C_1 \cos 5t + C_2 \sin 5t + \frac{16}{19}\cos 4.5t$

Step 4: Apply ICs:

$x(0) = C_1 + \frac{16}{19} = 0 \Rightarrow C_1 = -\frac{16}{19}$

$x'(t) = -5C_1 \sin 5t + 5C_2 \cos 5t - \frac{72}{19}\sin 4.5t$

$x'(0) = 5C_2 = 0 \Rightarrow C_2 = 0$

Final Answer:

$\boxed{x(t) = -\frac{16}{19}\cos 5t + \frac{16}{19}\cos 4.5t = \frac{16}{19}(\cos 4.5t - \cos 5t)}$