Ordinary Differential Equations – Book-Back Answer Key (Exercise 10.9)
Correct option for every multiple-choice question in Exercise 10.9, with a one-line justification. Each answer was solved independently; all twenty-five agree with the official textbook key.
| Q.No | Correct answer | Reason |
|---|---|---|
| Q1 | Option A \(2,\ 3\) | Highest derivative is 2nd order; cubing to clear the \(1/3\) power makes \( \frac{d^2y}{dx^2} \) appear to the 3rd power — order \(2\), degree \(3\). |
| Q2 | Option B \( \dfrac{d^2y}{dx^2} + y = 0 \) | Differentiating \( y=A\cos(x+B) \) twice gives \( y''=-y \), i.e. \( \frac{d^2y}{dx^2}+y=0 \). |
| Q3 | Option C \(1,\ 1\) | The square roots sit on the coefficients, not on \( \frac{dy}{dx} \); only the first derivative appears — order \(1\), degree \(1\). |
| Q4 | Option B \(3\) | A general circle has three free constants \( h,k,a \), so three differentiations are needed — order \(3\) (a common mis-answer is \(2\)). |
| Q5 | Option B \( \dfrac{d^2y}{dx^2} - y = 0 \) | \( y=Ae^{x}+Be^{-x} \) satisfies \( y''=y \), i.e. \( \frac{d^2y}{dx^2}-y=0 \). |
| Q6 | Option C \( y = kx \) | Separating \( \frac{dy}{y}=\frac{dx}{x} \) and integrating gives \( y=kx \). |
| Q7 | Option C parabola | Separating and integrating gives \( (y+3)^2=cx \), which is a parabola. |
| Q8 | Option B \( y = c\,e^{-\int p\,dx} \) | With \( Q=0 \) the linear solution collapses to \( y=c\,e^{-\int p\,dx} \). |
| Q9 | Option B \( \dfrac{e^{x}}{x} \) | After moving \( \frac{y}{x} \) left, \( P=1-\frac{1}{x} \), so I.F. \( =e^{\,x-\ln x}=\frac{e^{x}}{x} \). |
| Q10 | Option C \( \dfrac{1}{x} \) | \( e^{\int P\,dx}=x\Rightarrow\int P\,dx=\ln x\Rightarrow P=\frac{1}{x} \). |
| Q11 | Option C \(1\) | The series sums to \( e^{dy/dx} \), giving \( \frac{dy}{dx}=\ln y \) — degree \(1\). |
| Q12 | Option C \( p > q \) | Order \( p=2 \), degree \( q=1 \), so \( p>q \). |
| Q13 | Option A \( y + \sin^{-1}x = c \) | Integrating \( -\frac{dx}{\sqrt{1-x^{2}}} \) gives \( y+\sin^{-1}x=c \). |
| Q14 | Option A \( y = c\,e^{x^{2}} \) | Separating \( \frac{dy}{y}=2x\,dx \) gives \( y=c\,e^{x^{2}} \). |
| Q15 | Option B \( e^{x} + e^{-y} = c \) | \( \frac{dy}{dx}=e^{x}e^{y} \) separates and integrates to \( e^{x}+e^{-y}=c \). |
| Q16 | Option C \( \dfrac{1}{2^{x}} - \dfrac{1}{2^{y}} = c \) | \( 2^{-y}\,dy=2^{-x}\,dx \) integrates to \( \frac{1}{2^{x}}-\frac{1}{2^{y}}=c \). |
| Q17 | Option B \( \varphi\!\left(\dfrac{y}{x}\right) = kx \) | With \( y=vx \) it separates to \( \frac{\varphi'(v)}{\varphi(v)}dv=\frac{dx}{x} \), giving \( \varphi\!\left(\frac{y}{x}\right)=kx \). |
| Q18 | Option D \( \cot x \) | \( e^{\int P\,dx}=\sin x\Rightarrow P=\frac{d}{dx}\ln(\sin x)=\cot x \). |
| Q19 | Option B \( n,\ n+1 \) | A general solution of order \(m\) has \(m\) constants, so the counts are \( n \) and \( n+1 \). |
| Q20 | Option D \(0\) | A particular solution fixes every constant, leaving \(0\). |
| Q21 | Option A \( \dfrac{1}{x+1} \) | Linear form gives \( P=-\frac{1}{x+1} \), so I.F. \( =e^{-\ln(x+1)}=\frac{1}{x+1} \). |
| Q22 | Option A \( P = c\,e^{kt} \) | \( \frac{dP}{dt}=kP \) integrates to \( P=c\,e^{kt} \) (growth). |
| Q23 | Option B \( P = c\,e^{-kt} \) | Decay \( \frac{dP}{dt}=-kP \) integrates to \( P=c\,e^{-kt} \). |
| Q24 | Option B \(-2\) | Integrating and matching the \( x^{2} \) and \( y^{2} \) coefficients gives \( \frac{a}{2}=-1 \), so \( a=-2 \). |
| Q25 | Option A \( y = x^{3} + 2 \) | \( y=x^{3}+c \) through \( (-1,1) \) gives \( c=2 \), so \( y=x^{3}+2 \). |