Solved Examples

How to Use This Section

Twenty-five fully worked problems in five batches: order/degree and verification, separable equations, homogeneous equations, linear equations, and mixed curve/model problems at miscellaneous level.

The classification ritual for every solving problem:

1. Is the right side a product $g(x)h(y)$? → separable.
2. Is it a function of the ratio $\frac yx$ (uniform degrees)? → homogeneous: $y = vx$.
3. Do $y$ and $y'$ appear linearly? → linear: compute the I.F. (Check the mirror orientation too — linear in $x$.)
4. Solve generally, then apply any given condition to fix $C$.

Key Point: Marks are lost at classification, not at integration. Spend the first fifteen seconds of every problem naming its type — the rest is Chapter 7 machinery.

Batch 1 — Order, Degree and Verification (Easy)

Example 1: A clean third-order equation

Find the order and degree of $y''' + 2y'' + y' = 0$.

Solution:

1. Order: highest derivative $y'''$ — order $3$.
2. Degree: polynomial in all derivatives, power of $y'''$ is $1$.

Final Answer: order $3$, degree $1$.

Example 2: Clear the root first

Find the order and degree of $y'' = \sqrt{1 + \left(y'\right)^2}$.

Solution:

1. Clear the radical: square both sides: $\left(y''\right)^2 = 1 + \left(y'\right)^2$ — now polynomial in derivatives.
2. Read off: order $2$; power of $y''$ is $2$.

Final Answer: order $2$, degree $2$.

Takeaway: Roots and fractions of derivatives must be cleared before reading the degree — but a $\sin(y')$ can never be cleared; that stays "not defined".

Example 3: Verification with a product

Verify that $y = x\sin 3x$ is a solution of $\dfrac{d^2y}{dx^2} + 9y - 6\cos 3x = 0$.

Solution:

1. Differentiate twice: $y' = \sin 3x + 3x\cos 3x$; $y'' = 6\cos 3x - 9x\sin 3x$.
2. Substitute: L.H.S. $= (6\cos 3x - 9x\sin 3x) + 9x\sin 3x - 6\cos 3x = 0$. ✓

Final Answer: Verified.

Example 4: A two-parameter damped-oscillation family

Verify that $y = c_1e^{ax}\cos bx + c_2e^{ax}\sin bx$ ($c_1, c_2$ arbitrary) solves $\dfrac{d^2y}{dx^2} - 2a\dfrac{dy}{dx} + \left(a^2 + b^2\right)y = 0$.

Solution:

1. First derivative: $y' = e^{ax}\left[(bc_2 + ac_1)\cos bx + (ac_2 - bc_1)\sin bx\right]$.
2. Second derivative: differentiating again multiplies in another mix of $a$'s and $b$'s.
3. Substitute and collect: the coefficients of both $\cos bx$ and $\sin bx$ cancel identically, giving L.H.S. $= e^{ax}\cdot 0 = 0$. ✓

Final Answer: Verified — and with two arbitrary constants matching order $2$, this family is the general solution.

Example 5: Reading the constants

Without solving, state how many arbitrary constants appear in (i) the general solution of $y'' - 2ay' + \left(a^2 + b^2\right)y = 0$, and (ii) any particular solution of it.

Solution:

1. (i): order $2$ → general solution has $2$ arbitrary constants (visible as $c_1, c_2$ in Example 4).
2. (ii): particular solutions have $0$ arbitrary constants, by definition.

Final Answer: (i) $2$; (ii) $0$.

Batch 2 — Variables Separable (Easy-Medium)

Example 6: A one-line separation

Solve $\dfrac{dy}{dx} = \left(1 + y^2\right)e^x$.

Solution:

1. Separate: $\frac{dy}{1 + y^2} = e^x dx$.
2. Integrate: $\tan^{-1}y = e^x + C$.

Final Answer: $\tan^{-1}y = e^x + C$.

Example 7: A logarithmic disguise

Find the particular solution of $\log\left(\dfrac{dy}{dx}\right) = 3x + 4y$, given $y = 0$ when $x = 0$.

Solution:

1. Exponentiate: $\frac{dy}{dx} = e^{3x + 4y} = e^{3x}e^{4y}$ — separable.
2. Separate and integrate: $\int e^{-4y}dy = \int e^{3x}dx$ gives $\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C$, i.e. $4e^{3x} + 3e^{-4y} + 12C = 0$.
3. Apply $(0,0)$: $4 + 3 + 12C = 0$, so $12C = -7$.

Final Answer: $4e^{3x} + 3e^{-4y} - 7 = 0$.

Example 8: Exponential of the unknown

Solve $\dfrac{dy}{dx} = x^3e^{-y}$.

Solution:

1. Separate: $e^y dy = x^3dx$.
2. Integrate: $e^y = \frac{x^4}{4} + C$.

Final Answer: $e^y = \dfrac{x^4}{4} + C$, i.e. $y = \log\left(\dfrac{x^4}{4} + C\right)$.

Example 9: Bacterial growth

In a culture the bacteria count is $1{,}00{,}000$ and increases by 10% in 2 hours. If the growth rate is proportional to the count, in how many hours will the count reach $2{,}00{,}000$?

Solution:

1. Model: $\frac{dN}{dt} = kN$ gives $N = N_0e^{kt}$ with $N_0 = 1{,}00{,}000$.
2. Use the 10% data: $1{,}10{,}000 = N_0e^{2k}$, so $e^{2k} = \frac{11}{10}$, i.e. $k = \frac12\log\frac{11}{10}$.
3. Doubling: $2 = e^{kt}$ gives $t = \frac{\log 2}{k}$.

Final Answer: $t = \dfrac{2\log 2}{\log\frac{11}{10}}$ hours.

Example 10: The inflating balloon

A spherical balloon's volume changes at a constant rate. Its radius is $3$ units initially and $6$ units after $3$ seconds. Find the radius after $t$ seconds.

Solution:

1. Model: $\frac{dV}{dt} = k$ with $V = \frac43\pi r^3$, so $4\pi r^2\frac{dr}{dt} = k$.
2. Separate and integrate: $\frac{4}{3}\pi r^3 = kt + C$ — i.e. $r^3$ is linear in $t$: $r^3 = at + b$.
3. Fit the data: $r(0) = 3$ gives $b = 27$; $r(3) = 6$ gives $216 = 3a + 27$, so $a = 63$.

Final Answer: $r(t) = \left(63t + 27\right)^{1/3}$ units.

Batch 3 — Homogeneous Equations (Medium)

Example 11: Log-and-arctangent pair

Solve $(x - y)\,dy - (x + y)\,dx = 0$.

Solution:

1. Rearrange: $\frac{dy}{dx} = \frac{x + y}{x - y}$ — homogeneous (uniform degree 1).
2. Substitute $y = vx$: $x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v^2}{1 - v}$.
3. Separate: $\frac{1 - v}{1 + v^2}dv = \frac{dx}{x}$; integrate term-wise: $\tan^{-1}v - \frac12\log\left(1 + v^2\right) = \log|x| + C$.
4. Back-substitute and merge logs:

Final Answer: $\tan^{-1}\dfrac{y}{x} = \log\sqrt{x^2 + y^2} + C$.

Example 12: A grand trig-homogeneous equation

Solve $(x\,dy - y\,dx)\,y\sin\left(\dfrac{y}{x}\right) = (y\,dx + x\,dy)\,x\cos\left(\dfrac{y}{x}\right)$.

Solution:

1. Collect $dy$ and $dx$: $\frac{dy}{dx} = \frac{xy\cos\frac yx + y^2\sin\frac yx}{xy\sin\frac yx - x^2\cos\frac yx}$ — dividing by $x^2$ shows a pure function of $\frac yx$: homogeneous.
2. Substitute $y = vx$: $x\frac{dv}{dx} = \frac{2v\cos v}{v\sin v - \cos v}$.
3. Separate: $\frac{v\sin v - \cos v}{v\cos v}dv = \frac{2\,dx}{x}$, i.e. $\int\tan v\,dv - \int\frac{dv}{v} = 2\int\frac{dx}{x}$.
4. Integrate and merge: $\log\left|\frac{\sec v}{v}\right| = \log\left|x^2\right| + \log|C_1|$, so $\frac{\sec v}{vx^2} = \pm C_1$.
5. Back-substitute $v = \frac yx$:

Final Answer: $\sec\left(\dfrac{y}{x}\right) = Cxy$.

Example 13: A sine-driven equation

Solve $x\dfrac{dy}{dx} - y + x\sin\left(\dfrac{y}{x}\right) = 0$.

Solution:

1. Rearrange: $\frac{dy}{dx} = \frac yx - \sin\frac yx$ — homogeneous.
2. Substitute $y = vx$: $x\frac{dv}{dx} = -\sin v$.
3. Separate: $\mathrm{cosec}\,v\,dv = -\frac{dx}{x}$; integrate: $\log\left|\mathrm{cosec}\,v - \cot v\right| = -\log|x| + \log C$.
4. Simplify ($\mathrm{cosec}\,v - \cot v = \frac{1 - \cos v}{\sin v}$) and back-substitute:

Final Answer: $x\left(1 - \cos\dfrac{y}{x}\right) = C\sin\dfrac{y}{x}$.

Example 14: Particular solution with a log

Solve $2xy + y^2 - 2x^2\dfrac{dy}{dx} = 0$, given $y = 2$ when $x = 1$.

Solution:

1. Rearrange: $\frac{dy}{dx} = \frac{2xy + y^2}{2x^2} = v + \frac{v^2}{2}$ with $v = \frac yx$.
2. Substitute: $x\frac{dv}{dx} = \frac{v^2}{2}$; separate: $\frac{2\,dv}{v^2} = \frac{dx}{x}$.
3. Integrate: $-\frac2v = \log|x| + C$, i.e. $-\frac{2x}{y} = \log|x| + C$.
4. Apply $(1, 2)$: $-1 = 0 + C$, so $C = -1$: $\frac{2x}{y} = 1 - \log|x|$.

Final Answer: $y = \dfrac{2x}{1 - \log|x|}$ (for $x \neq e$).

Example 15: A squared-sine equation

Solve $\left[x\sin^2\left(\dfrac{y}{x}\right) - y\right]dx + x\,dy = 0$, given $y = \dfrac{\pi}{4}$ when $x = 1$.

Solution:

1. Rearrange: $\frac{dy}{dx} = \frac yx - \sin^2\frac yx$ — homogeneous.
2. Substitute $y = vx$: $x\frac{dv}{dx} = -\sin^2v$; separate: $\mathrm{cosec}^2v\,dv = -\frac{dx}{x}$.
3. Integrate: $-\cot v = -\log|x| - C$, i.e. $\cot v = \log|x| + C$.
4. Apply the condition: $\cot\frac{\pi}{4} = 0 + C$ gives $C = 1$.

Final Answer: $\cot\left(\dfrac{y}{x}\right) = \log|x| + 1 = \log|ex|$.

Batch 4 — Linear Equations (Medium)

Example 16: Constant coefficient, trig right side

Solve $\dfrac{dy}{dx} + 2y = \sin x$.

Solution:

1. I.F.: $e^{2x}$.
2. Solve: $ye^{2x} = \int e^{2x}\sin x\,dx + C$ — the loop-back by parts gives $\frac{e^{2x}(2\sin x - \cos x)}{5}$.
3. Divide:

Final Answer: $y = \dfrac{2\sin x - \cos x}{5} + Ce^{-2x}$.

Example 17: Secant coefficient

Solve $\dfrac{dy}{dx} + (\sec x)\,y = \tan x$ (for $0 \leq x < \frac{\pi}{2}$).

Solution:

1. I.F.: $e^{\int\sec x\,dx} = e^{\log|\sec x + \tan x|} = \sec x + \tan x$.
2. Solve: $y(\sec x + \tan x) = \int\tan x(\sec x + \tan x)dx + C = \int\left(\sec x\tan x + \sec^2x - 1\right)dx + C$.
3. Integrate: $\sec x + \tan x - x + C$.

Final Answer: $y\left(\sec x + \tan x\right) = \sec x + \tan x - x + C$.

Example 18: A ready-made left side

Solve $\left(1 + x^2\right)dy + 2xy\,dx = \cot x\,dx$ (with $x \neq 0$).

Solution:

1. Standardise: $\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{\cot x}{1 + x^2}$: $P = \frac{2x}{1+x^2}$.
2. I.F.: $e^{\log(1+x^2)} = 1 + x^2$.
3. Solve: $y\left(1 + x^2\right) = \int\cot x\,dx + C = \log|\sin x| + C$.

Final Answer: $y\left(1 + x^2\right) = \log|\sin x| + C$.

Takeaway: The original equation was already $d\left[y(1+x^2)\right] = \cot x\,dx$ — recognising exact left sides skips the whole I.F. computation.

Example 19: Linear in $x$ with an inverse-trig drive

Solve $\left(\tan^{-1}y - x\right)dy = \left(1 + y^2\right)dx$.

Solution:

1. Reorient: $\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1}y}{1 + y^2}$ — linear in $x$ with $P_1 = \frac{1}{1+y^2}$.
2. I.F.: $e^{\tan^{-1}y}$.
3. Solve (substitute $t = \tan^{-1}y$ in the integral): $xe^{\tan^{-1}y} = \int te^t dt + C = e^t(t - 1) + C$.
4. Divide:

Final Answer: $x = \left(\tan^{-1}y - 1\right) + Ce^{-\tan^{-1}y}$.

Example 20: Particular solution with matching arctangents

Solve $\left(1 + x^2\right)\dfrac{dy}{dx} + 2xy = \dfrac{1}{1 + x^2}$, given $y = 0$ when $x = 1$.

Solution:

1. Standardise: $P = \frac{2x}{1+x^2}$; I.F. $= 1 + x^2$.
2. Solve: $y\left(1 + x^2\right) = \int\frac{dx}{1 + x^2} + C = \tan^{-1}x + C$.
3. Apply $(1, 0)$: $0 = \frac{\pi}{4} + C$, so $C = -\frac{\pi}{4}$.

Final Answer: $y\left(1 + x^2\right) = \tan^{-1}x - \dfrac{\pi}{4}$.

Batch 5 — Curves and Mixed Problems (Medium-Hard)

Example 21: Curve through the origin

Find the equation of the curve through the origin whose tangent slope at $(x, y)$ equals the sum of the coordinates.

Solution:

1. Translate: $\frac{dy}{dx} = x + y$, i.e. $\frac{dy}{dx} - y = x$ — linear, I.F. $= e^{-x}$.
2. Solve: $ye^{-x} = \int xe^{-x}dx + C = -(x + 1)e^{-x} + C$, so $y = -(x + 1) + Ce^x$.
3. Apply $(0, 0)$: $0 = -1 + C$, so $C = 1$.

Final Answer: $y = e^x - x - 1$, i.e. $x + y + 1 = e^x$.

Example 22: A worded slope condition

Find the equation of the curve through $(0, 2)$, given that the sum of the coordinates of any point exceeds the magnitude of the tangent slope there by $5$.

Solution:

1. Translate: $x + y = \frac{dy}{dx} + 5$, i.e. $\frac{dy}{dx} - y = x - 5$ — linear, I.F. $= e^{-x}$.
2. Solve: $ye^{-x} = \int(x - 5)e^{-x}dx + C = -(x - 4)e^{-x} + C$, so $y = 4 - x + Ce^x$.
3. Apply $(0, 2)$: $2 = 4 + C$, so $C = -2$.

Final Answer: $y = 4 - x - 2e^x$.

Example 23: A product-rule left side

Solve $x\dfrac{dy}{dx} + y = x\log x$ (for $x > 0$).

Solution:

1. Recognise the exact form: the left side is $\frac{d}{dx}(xy)$.
2. Integrate: $xy = \int x\log x\,dx + C$ — by parts: $\frac{x^2}{2}\log x - \frac{x^2}{4} + C$.

Final Answer: $xy = \dfrac{x^2}{2}\log x - \dfrac{x^2}{4} + C$.

Example 24: A cubic-sine particular solution

Solve $\dfrac{dy}{dx} - 3y\cot x = \sin 2x$, given $y = 2$ when $x = \dfrac{\pi}{2}$.

Solution:

1. I.F.: $e^{-3\int\cot x\,dx} = e^{-3\log\sin x} = \dfrac{1}{\sin^3x}$.
2. Solve: $\frac{y}{\sin^3x} = \int\frac{2\sin x\cos x}{\sin^3x}dx + C = 2\int\cot x\,\mathrm{cosec}\,x\,dx + C = -2\,\mathrm{cosec}\,x + C$.
3. Apply the condition: at $x = \frac{\pi}{2}$: $2 = -2 + C$, so $C = 4$.
4. Isolate: $y = 4\sin^3x - 2\sin^2x$.

Final Answer: $y = 4\sin^3 x - 2\sin^2 x$.

Example 25: A one-line integrating factor classic

Solve $\dfrac{dy}{dx} + y\tan x = \sec x$.

Solution:

1. I.F.: $e^{\int\tan x\,dx} = e^{\log\sec x} = \sec x$.
2. Solve: $y\sec x = \int\sec^2x\,dx + C = \tan x + C$.
3. Divide:

Final Answer: $y = \sin x + C\cos x$.

Takeaway: With the four-entry I.F. speed-table ($\tan x \to \sec x$, $\cot x \to \sin x$, $\frac ax \to x^a$, constant $k \to e^{kx}$), problems like this compress to three written lines — full marks, minimal ink.