Chapter 9: Differential Equations

Dear Class 12 Student! If derivatives represent the "rate of change", then Differential Equations represent the laws of nature themselves. From predicting population growth to calculating how fast a cup of coffee cools down, everything is modeled using DEs. For the Boards, identifying the Order & Degree, and solving Linear Differential Equations (LDE) are absolute 5-mark guarantees. For JEE, equations reducible to variable separable and orthogonal trajectories are game-changers. Let's solve!

1. Introduction, Order, and Degree (1-Mark Board Guarantee)

Differential Equation (DE): An equation that involves independent variables (e.g., $x$), dependent variables (e.g., $y$), and the derivatives of the dependent variables with respect to the independent variables (e.g., $\frac{dy}{dx}, y', y''$).

Order and Degree Order of a DE: The order of the highest order derivative appearing in the equation. (The order is ALWAYS uniquely defined).

Degree of a DE: The highest power (positive integer only) raised to the highest order derivative, provided the differential equation is a polynomial equation in its derivatives.

Crucial Rule for Degree: The DE must be completely free from radicals (roots) and fractional powers concerning the derivatives. You must mathematically clear them before finding the degree.

When is Degree "Not Defined"? If a derivative is forever trapped inside a transcendental function (e.g., $\sin(\frac{dy}{dx})$, $e^{y'}$, or $\log(y'')$), the equation cannot be written as a polynomial of derivatives. In such cases, the degree is strictly Not Defined.

Practice Problem 1: Order & Degree Question: Find the order and degree of the following differential equations:
(i) $\left( \frac{d^2y}{dx^2} \right)^3 + \left( \frac{dy}{dx} \right)^4 + y = 0$
(ii) $y'' + (y')^2 + \sin(y') = 0$
(iii) $\left[ 1 + \left(\frac{dy}{dx}\right)^2 \right]^{3/2} = k \frac{d^2y}{dx^2}$

Solution:
(i) The highest derivative is $\frac{d^2y}{dx^2}$ (2nd order). The power raised to this highest derivative is 3.
$\implies$ Order = 2, Degree = 3.

(ii) The highest derivative is $y''$ (2nd order). However, $y'$ is trapped inside the sine function ($\sin(y')$). It is not a polynomial equation in derivatives.
$\implies$ Order = 2, Degree = Not Defined.

(iii) We must remove the fractional power ($3/2$) by squaring both sides:
$\left[ 1 + \left(\frac{dy}{dx}\right)^2 \right]^3 = k^2 \left( \frac{d^2y}{dx^2} \right)^2$
The highest derivative is $\frac{d^2y}{dx^2}$ (2nd order). Its power is 2.
$\implies$ Order = 2, Degree = 2.

2. General and Particular Solutions

Unlike normal algebraic equations where solutions are numbers, the solution to a Differential Equation is a Function (a curve).

Family of Curves (General Solution)

The General Solution $y = C e^x$ represents an infinite family of curves. A Particular Solution (e.g., $y = 2e^x$) picks exactly one specific curve from this family based on a given boundary condition.

General Solution: A solution that contains exactly as many arbitrary constants (e.g., $A, B, C$) as the order of the differential equation. Geometrically, it represents a family of curves.
Particular Solution: A solution obtained from the general solution by giving specific numerical values to the arbitrary constants. These values are found by plugging in given initial boundary conditions (e.g., "$y=1$ when $x=0$"). It represents one specific curve.

Practice Problem 2: Verifying a Solution Question: Verify that $y = A \sin x + B \cos x$ is a general solution to the differential equation $\frac{d^2y}{dx^2} + y = 0$.

Solution:
1. Given: $y = A \sin x + B \cos x$.
2. Differentiate w.r.t $x$:
$\frac{dy}{dx} = A \cos x - B \sin x$.
3. Differentiate again w.r.t $x$ to find the 2nd order derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(A \cos x - B \sin x) = A(-\sin x) - B(\cos x)$.
$\frac{d^2y}{dx^2} = -(A \sin x + B \cos x)$.
4. Substitute the expression for $y$ back into the RHS:
$\frac{d^2y}{dx^2} = -y \implies \mathbf{\frac{d^2y}{dx^2} + y = 0}$.
5. Since it satisfies the DE perfectly, it is indeed a solution. Because it has two arbitrary constants ($A, B$) for a 2nd order DE, it is the General Solution.

3. Formation of a Differential Equation

Note: Sometimes rationalized in recent NCERT syllabi, but absolutely essential for setting up R.D. Sharma word problems.

Concept: If you are given a geometric equation representing a family of curves with $n$ arbitrary constants, your goal is to form a differential equation that is completely free of those arbitrary constants.

Algorithm for Formation 1. Count: Count the number of arbitrary constants ($n$) in the given equation.
2. Differentiate: Differentiate the given equation exactly $n$ times to obtain $n$ new derivative equations.
3. Eliminate: Use algebraic manipulation among the original equation and the $n$ differentiated equations to permanently eliminate all arbitrary constants.
4. The resulting equation is the required DE. (Its order will be exactly $n$).

Practice Problem 3: Formation Question: Form the differential equation representing the family of circles having their center at the origin.

Solution:
1. The general equation of a family of circles centered at the origin is: $x^2 + y^2 = r^2$.
2. Here, there is only one arbitrary constant ($r$). So, we will differentiate exactly once.
3. Differentiate w.r.t $x$:
$2x + 2y \frac{dy}{dx} = 0$
4. Divide by 2 to simplify:
$x + y \frac{dy}{dx} = 0 \implies \mathbf{x + y y' = 0}$.
5. The arbitrary constant $r$ has been naturally eliminated. This is the required Differential Equation.

4. Methods of Solving First Order, First Degree DEs

This is the core of the chapter. Every question you encounter must be categorized into one of three methods before solving.

Method A: Variable Separable Form

If the equation can be algebraically manipulated so that all $x$ terms and $dx$ are on one side, and all $y$ terms and $dy$ are on the other.

Form: $f(x) dx = g(y) dy$
Algorithm: Separate the variables, then simply integrate both sides: $\int f(x) dx = \int g(y) dy + C$.

Practice Problem 4: Standard Variable Separable Question: Solve the differential equation $\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}$.

Solution:
1. Use exponent rules to expand: $\frac{dy}{dx} = e^x \cdot e^{-y} + x^2 \cdot e^{-y}$.
2. Factor out $e^{-y}$ on the RHS: $\frac{dy}{dx} = e^{-y} (e^x + x^2)$.
3. Separate variables (move $y$ terms to left, $x$ to right):
$\frac{dy}{e^{-y}} = (e^x + x^2) dx \implies e^y dy = (e^x + x^2) dx$.
4. Integrate both sides:
$\int e^y dy = \int (e^x + x^2) dx$
5. $e^y = e^x + \frac{x^3}{3} + C$ (General Solution).

R.D. Sharma / JEE Focus: Equations Reducible to Variable Separable When you see a linear expression of $x$ and $y$ trapped inside a function, e.g., $\frac{dy}{dx} = f(ax + by + c)$.
Algorithm:
1. Substitute the trapped expression: Let $ax + by + c = t$.
2. Differentiate w.r.t $x$: $a + b \frac{dy}{dx} = \frac{dt}{dx} \implies \frac{dy}{dx} = \frac{1}{b} \left( \frac{dt}{dx} - a \right)$.
3. Substitute back into the DE. It will now be purely Variable Separable in terms of $t$ and $x$.

Practice Problem 5: Reducible to Variable Separable Question: Solve the differential equation $\frac{dy}{dx} = (x+y)^2$.

Solution:
1. We cannot separate $x$ and $y$ algebraically because they are squared together. Let $x + y = t$.
2. Differentiate w.r.t $x$: $1 + \frac{dy}{dx} = \frac{dt}{dx} \implies \frac{dy}{dx} = \frac{dt}{dx} - 1$.
3. Substitute into original DE:
$\frac{dt}{dx} - 1 = t^2 \implies \frac{dt}{dx} = t^2 + 1$.
4. Now it is Variable Separable! Separate $t$ and $x$:
$\frac{dt}{t^2 + 1} = dx$.
5. Integrate both sides:
$\int \frac{dt}{t^2 + 1} = \int dx \implies \tan^{-1}(t) = x + C$.
6. Replace $t$ with $x+y$:
$\tan^{-1}(x+y) = x + C$.

Method B: Homogeneous Differential Equations

Homogeneous Function: A function $F(x, y)$ is homogeneous of degree $n$ if substituting $x \rightarrow \lambda x$ and $y \rightarrow \lambda y$ yields $F(\lambda x, \lambda y) = \lambda^n F(x, y)$. Simply put, every term has the exact same total algebraic degree.

Algorithm: Homogeneous DE Form 1: $\frac{dy}{dx} = \frac{f(x,y)}{g(x,y)}$ (where $f$ and $g$ are homogeneous of same degree).
1. Substitute $y = vx$.
2. Differentiate using product rule: $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
3. Substitute these into the DE. The variable $x$ will magically cancel out from the RHS fraction.
4. The equation reduces to Variable Separable form in terms of $v$ and $x$. Integrate.
5. Final step: Replace $v$ with $y/x$.

Form 2 (Special Case): If the equation has terms like $x/y$ or $e^{x/y}$, isolate $\frac{dx}{dy}$.
Substitute $x = vy \implies \frac{dx}{dy} = v + y \frac{dv}{dy}$.

Practice Problem 6: Homogeneous DE Question: Solve $(x^2 + y^2)dx - 2xy \, dy = 0$.

Solution:
1. Isolate $dy/dx$: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$. (Every term is degree 2. It is homogeneous).
2. Put $y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx}$.
3. Substitute: $v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2(1 + v^2)}{2x^2 v} = \frac{1 + v^2}{2v}$.
4. Isolate $x \frac{dv}{dx}$ term:
$x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$.
5. Separate variables:
$\frac{2v}{1 - v^2} dv = \frac{dx}{x}$.
6. Integrate both sides. Let $1 - v^2 = t \implies -2v dv = dt$.
$\int \frac{-dt}{t} = \int \frac{dx}{x} \implies -\log|1 - v^2| = \log|x| + C$.
7. Put $v = y/x$:
$-\log|1 - \frac{y^2}{x^2}| = \log|x| + C \implies \log|x| + \log|\frac{x^2 - y^2}{x^2}| = -C = \log K$.
$\log\left| x \cdot \frac{x^2 - y^2}{x^2} \right| = \log K \implies \mathbf{\frac{x^2 - y^2}{x} = K \implies x^2 - y^2 = Kx}$.

Method C: Linear Differential Equations (LDE) - High Board Priority

Algorithm: Type 1 LDE Standard Form: $\frac{dy}{dx} + P y = Q$
Condition: $P$ and $Q$ must be functions of $x$ only, or constants.
1. Find the Integrating Factor (I.F.): $I.F. = e^{\int P \, dx}$.
2. Write the direct solution formula:
$$ y \times (I.F.) = \int (Q \times I.F.) \, dx + C $$ 3. Integrate the RHS to finish.

Type 2 (Reverse): $\frac{dx}{dy} + P_1 x = Q_1$ (Use when there's a single $x$ term but multiple complex $y$ terms).
$I.F. = e^{\int P_1 \, dy}$. Solution: $x \times (I.F.) = \int (Q_1 \times I.F.) \, dy + C$.

Practice Problem 7: Standard LDE Question: Solve the DE: $x \frac{dy}{dx} + 2y = x^2 \quad (x \neq 0)$.

Solution:
1. Make the coefficient of $dy/dx$ unity by dividing entirely by $x$:
$\frac{dy}{dx} + \frac{2}{x} y = x$.
2. Compare with $\frac{dy}{dx} + Py = Q$. Here $P = \frac{2}{x}$ and $Q = x$.
3. Find Integrating Factor (I.F.):
$I.F. = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2\log x} = e^{\log(x^2)} = \mathbf{x^2}$.
4. Write the solution formula:
$y \times (I.F.) = \int (Q \times I.F.) \, dx + C$
$y \cdot x^2 = \int (x \cdot x^2) \, dx + C$
$y x^2 = \int x^3 \, dx + C$
5. Integrate: $y x^2 = \frac{x^4}{4} + C \implies \mathbf{y = \frac{x^2}{4} + cx^{-2}}$.

5. Equations Reducible to Linear Form (Bernoulli's Equation)

Note: Strictly R.D. Sharma / JEE Mains Level. Very highly tested in competitive exams.

Form: $\frac{dy}{dx} + P y = Q y^n$ (where $n \neq 0, 1$).

Algorithm: Bernoulli's Equation 1. Divide the entire equation by $y^n$ to get: $y^{-n} \frac{dy}{dx} + P y^{1-n} = Q$.
2. Substitute the middle term: Let $y^{1-n} = t$.
3. Differentiate w.r.t $x$: $(1-n) y^{-n} \frac{dy}{dx} = \frac{dt}{dx} \implies y^{-n} \frac{dy}{dx} = \frac{1}{1-n} \frac{dt}{dx}$.
4. Substitute back to convert the DE into a standard LDE in terms of $t$ and $x$:
$\frac{1}{1-n} \frac{dt}{dx} + P t = Q \implies \mathbf{\frac{dt}{dx} + (1-n)P \cdot t = (1-n)Q}$.
5. Solve as a standard LDE and replace $t$ at the end.

Practice Problem 8: Bernoulli's Equation Question: Solve the DE: $\frac{dy}{dx} + \frac{y}{x} = y^2$.

Solution:
1. This is a Bernoulli equation with $n=2$. Divide by $y^2$:
$y^{-2} \frac{dy}{dx} + \frac{1}{x} y^{-1} = 1$
2. Let $y^{-1} = t$. Differentiating w.r.t $x$: $-y^{-2} \frac{dy}{dx} = \frac{dt}{dx} \implies y^{-2} \frac{dy}{dx} = -\frac{dt}{dx}$.
3. Substitute into the equation:
$-\frac{dt}{dx} + \frac{1}{x} t = 1 \implies \frac{dt}{dx} - \frac{1}{x} t = -1$.
4. This is a standard LDE where $P = -1/x$ and $Q = -1$.
$I.F. = e^{\int -1/x dx} = e^{-\log x} = e^{\log(x^{-1})} = \mathbf{x^{-1} = 1/x}$.
5. Solution: $t \times (I.F.) = \int (Q \times I.F.) dx + C$
$t \cdot \frac{1}{x} = \int -1 \cdot \frac{1}{x} dx + C \implies \frac{t}{x} = -\log|x| + C$.
6. Replace $t = 1/y$:
$\frac{1}{xy} = -\log|x| + C \implies \mathbf{1 = xy(C - \log|x|)}$.

6. Applications of Differential Equations (Word Problems)

A. Population Growth / Radioactive Decay

The rate of change of population is directly proportional to the current population.

$\frac{dP}{dt} \propto P \implies \frac{dP}{dt} = kP$ (Variable Separable form).
Integrating yields: $P = P_0 e^{kt}$.

Practice Problem 9: Population Growth Question: In a bank, principal increases continuously at the rate of 5% per year. Find the time required for Rs 1000 to double itself. (Given $\ln 2 = 0.693$).

Solution:
1. Let $P$ be the principal at time $t$. The rate of change is 5% of $P$.
$\frac{dP}{dt} = \frac{5}{100} P = 0.05 P$.
2. Separate variables: $\frac{dP}{P} = 0.05 dt$.
3. Integrate: $\int \frac{dP}{P} = \int 0.05 dt \implies \ln P = 0.05t + C$.
4. At $t=0, P=1000 \implies \ln(1000) = C$.
Equation is: $\ln P = 0.05t + \ln 1000 \implies \ln(P/1000) = 0.05t$.
5. To double, $P = 2000$. Find $t$:
$\ln(2000/1000) = 0.05t \implies \ln 2 = 0.05t \implies 0.693 = 0.05t$.
$t = \frac{0.693}{0.05} = \frac{69.3}{5} = \mathbf{13.86 \text{ years}}$.

JEE Advanced Focus: Orthogonal Trajectories Concept: Find a family of curves that intersects a given family of curves perfectly at right angles.
Algorithm:
1. Differentiate the given family to form its DE, eliminating arbitrary constants. (Let this DE be $dy/dx = f(x,y)$).
2. Since the required family cuts orthogonally, $m_1 m_2 = -1$.
Crucial Step: Replace $\frac{dy}{dx}$ with $\mathbf{-\frac{dx}{dy}}$ in the DE.
3. Solve this new Differential Equation using standard methods. The solution is the family of orthogonal trajectories.

Practice Problem 10: Orthogonal Trajectories Question: Find the orthogonal trajectories of the family of parabolas $y^2 = cx$.

Solution:
1. Form DE: Differentiate $y^2 = cx$ w.r.t $x$: $2y \frac{dy}{dx} = c$.
Eliminate $c$ using the original equation $c = y^2/x$:
$2y \frac{dy}{dx} = \frac{y^2}{x} \implies 2x \frac{dy}{dx} = y$. (This is the DE of the parabolas).
2. Orthogonal Swap: Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ to get the DE of the trajectories.
$2x \left( -\frac{dx}{dy} \right) = y \implies -2x dx = y dy$.
3. Solve new DE: Integrate both sides.
$\int y dy = \int -2x dx \implies \frac{y^2}{2} = -x^2 + k$.
Multiply by 2: $y^2 = -2x^2 + 2k \implies \mathbf{2x^2 + y^2 = K}$.
4. Conclusion: The orthogonal trajectories form a family of Ellipses.