Differential Equations: Miscellaneous Exercise Class 12 NCERT

Diagnosing Differential Equations: The First Challenge

In the differential equations miscellaneous ex class 12 ncert, the biggest hurdle is not actually performing the integration, but identifying which technique to use. There are three primary types of first-order differential equations in the CBSE syllabus:

1. Variable Separable Form: If you can isolate all $y$ terms with $dy$ on one side and all $x$ terms with $dx$ on the other, i.e., $f(y)dy = g(x)dx$, integrate directly.
2. Homogeneous Equations: If the equation can be written as $\frac{dy}{dx} = F(x, y)$, where $F(\lambda x, \lambda y) = \lambda^0 F(x,y)$, it is homogeneous of degree 0. Use the substitution $y = vx$ (which yields $\frac{dy}{dx} = v + x\frac{dv}{dx}$).
3. Linear Differential Equations (LDE): If the equation fits the standard form $\frac{dy}{dx} + Py = Q$ (where $P$ and $Q$ are functions of $x$ only), find the Integrating Factor $I.F. = e^{\int P dx}$ and solve using $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + C$.

Step-by-Step Diagnostic & Solution Strategy

1. Step 1: Test for Variable Separable — Examine the terms. Can you factor out the variables? If you can group them into $h(y)dy = g(x)dx$, do so. This is always the fastest and least error-prone method.
2. Step 2: Test for Homogeneity — If variables cannot be separated easily, replace $x$ with $\lambda x$ and $y$ with $\lambda y$. If the expression remains algebraically unchanged, substitute $y = vx$ to transform it into a separable form in $v$ and $x$.
3. Step 3: Test for First-Order Linearity — If it is not homogeneous, rearrange it to check for LDE. Look out for single-power occurrences of $y$ (yielding $\frac{dy}{dx} + Py = Q$) or single-power occurrences of $x$ (yielding $\frac{dx}{dy} + P_1 x = Q_1$).
4. Step 4: Perform Integration and Apply Limits — Solve the resulting integrals. If the question asks for a 'particular solution', use the given initial boundary values $(x_0, y_0)$ to solve for the arbitrary constant $C$.

Rigorous Worked Examples

• Example 1 (Homogeneous): Solve the differential equation $(x^2 - y^2)dx + 2xy dy = 0$. Step 1: Express the equation in derivative form: $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ Step 2: Let $y = vx$, which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these values: $v + x\frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2v x^2} = \frac{v^2 - 1}{2v}$ Step 3: Isolate $v$ terms on one side: $x\frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = \frac{-(v^2 + 1)}{2v}$ Step 4: Separate variables and integrate: $\int \frac{2v}{v^2 + 1} dv = -\int \frac{1}{x} dx$ $\log(v^2 + 1) = -\log|x| + \log C \implies \log|x(v^2 + 1)| = \log C$ $x(v^2 + 1) = C$ Step 5: Substitute $v = y/x$ back: $x\left(\frac{y^2}{x^2} + 1\right) = C \implies x^2 + y^2 = Cx$.
• Example 2 (Linear in x): Find the general solution of the differential equation $(1 + y^2)dx + (x - e^{-\tan^{-1}y})dy = 0$. Step 1: Try to express as $\frac{dy}{dx}$. We get $\frac{dy}{dx} = \frac{-(1+y^2)}{x - e^{-\tan^{-1}y}}$. This is non-linear in $y$. Let us write it as $\frac{dx}{dy}$ instead: $\frac{dx}{dy} = \frac{-(x - e^{-\tan^{-1}y})}{1 + y^2} \implies \frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{e^{-\tan^{-1}y}}{1 + y^2}$ Step 2: This is a linear differential equation of the form $\frac{dx}{dy} + P_1 x = Q_1$, where $P_1 = \frac{1}{1 + y^2}$ and $Q_1 = \frac{e^{-\tan^{-1}y}}{1 + y^2}$. Step 3: Find the Integrating Factor (I.F.): $I.F. = e^{\int P_1 dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1}y}$ Step 4: Write the general solution: $x \cdot (I.F.) = \int (Q_1 \cdot I.F.) dy + C$ $x \cdot e^{\tan^{-1}y} = \int \left(\frac{e^{-\tan^{-1}y}}{1 + y^2} \cdot e^{\tan^{-1}y}\right) dy + C$ $x \cdot e^{\tan^{-1}y} = \int \frac{1}{1+y^2} dy + C \implies x \cdot e^{\tan^{-1}y} = \tan^{-1}y + C$.

Crucial Board Exam Tips & Common Mistakes

• The Linear-in-X Trap: Many students try to force every linear differential equation into the $\frac{dy}{dx} + Py = Q$ mould. If you find $y^2$ or $\sin y$ terms preventing this, check if the equation can be rearranged into $\frac{dx}{dy} + P_1 x = Q_1$.
• Sign of P: In $\frac{dy}{dx} + Py = Q$, if the equation is $\frac{dy}{dx} - \frac{2}{x}y = x^2$, the coefficient $P$ is $-\frac{2}{x}$ and not $\frac{2}{x}$. Forgetting the negative sign will completely ruin your Integrating Factor calculation.
• Arbitrary Constants: In homogeneous equations, using $\log C$ instead of $C$ as the constant of integration when integrating logarithmic terms simplifies algebraic fractions immensely. Always remember to state the constant!

Practice Questions with Solutions

• Q: Solve the differential equation: $y dx + (x - y^3) dy = 0$. A: Step 1: Rearrange the equation to find its type. Division by $dy$ yields: $y \frac{dx}{dy} + x - y^3 = 0 \implies \frac{dx}{dy} + \frac{1}{y} x = y^2$ Step 2: This is a linear differential equation of the form $\frac{dx}{dy} + P_1 x = Q_1$, where $P_1 = \frac{1}{y}$ and $Q_1 = y^2$. Step 3: Calculate the Integrating Factor: $I.F. = e^{\int \frac{1}{y} dy} = e^{\log y} = y$ Step 4: Write the general solution formula: $x \cdot (I.F.) = \int (Q_1 \cdot I.F.) dy + C \implies x \cdot y = \int (y^2 \cdot y) dy + C$ $xy = \int y^3 dy + C \implies xy = \frac{y^4}{4} + C$ Final answer: $xy = \frac{y^4}{4} + C$
• Q: Find the particular solution of the differential equation $\frac{dy}{dx} - 3y\cot x = \sin 2x$ given that $y = 2$ when $x = \frac{\pi}{2}$. A: Step 1: Identify standard form: $\frac{dy}{dx} + Py = Q$. Here, $P = -3\cot x$ and $Q = \sin 2x$. Step 2: Calculate the Integrating Factor: $I.F. = e^{\int -3\cot x dx} = e^{-3\log|\sin x|} = e^{\log(\sin x)^{-3}} = \frac{1}{\sin^3 x}$ Step 3: Write the general solution: $y \cdot \frac{1}{\sin^3 x} = \int \sin 2x \cdot \frac{1}{\sin^3 x} dx + C$ Step 4: Integrate the RHS using $\sin 2x = 2\sin x\cos x$: $\int \frac{2\sin x\cos x}{\sin^3 x} dx = 2 \int \frac{\cos x}{\sin^2 x} dx = -\frac{2}{\sin x}$ So, $\frac{y}{\sin^3 x} = -\frac{2}{\sin x} + C \implies y = -2\sin^2 x + C\sin^3 x$ Step 5: Apply boundary condition $y = 2$ when $x = \frac{\pi}{2}$: $2 = -2\sin^2(\pi/2) + C\sin^3(\pi/2) \implies 2 = -2(1) + C(1) \implies C = 4$ Final answer: $y = 4\sin^3 x - 2\sin^2 x$
• Q: Solve the differential equation: $\frac{dy}{dx} + y\tan x = x\cos x$. A: Step 1: Identify the form. This is a first-order LDE with $P = \tan x$ and $Q = x\cos x$. Step 2: Calculate the Integrating Factor: $I.F. = e^{\int \tan x dx} = e^{\log|\sec x|} = \sec x$ Step 3: Solve the equation: $y \cdot \sec x = \int (x\cos x \cdot \sec x) dx + C$ Step 4: Simplify the integral using $\cos x \cdot \sec x = 1$: $y \sec x = \int x dx + C \implies y\sec x = \frac{x^2}{2} + C$ Step 5: Multiply both sides by $\cos x$: $y = \left(\frac{x^2}{2} + C\right)\cos x$ Final answer: $y = \left(\frac{x^2}{2} + C\right)\cos x$
• Q: Show that the differential equation $(x - y)dy - (x + y)dx = 0$ is homogeneous and solve it. A: Step 1: Rewrite in standard derivative form: $\frac{dy}{dx} = \frac{x + y}{x - y}$ Step 2: Show homogeneity. Let $F(x,y) = \frac{x+y}{x-y}$. Then $F(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x - \lambda y} = \frac{\lambda(x+y)}{\lambda(x-y)} = F(x,y)$. Thus, it is homogeneous of degree 0. Step 3: Substitute $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$: $v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1+v}{1-v}$ Step 4: Isolate $v$ terms: $x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1 + v - v + v^2}{1-v} = \frac{1+v^2}{1-v}$ Step 5: Separate variables and integrate: $\int \frac{1-v}{1+v^2} dv = \int \frac{1}{x} dx \implies \int \frac{1}{1+v^2} dv - \frac{1}{2}\int \frac{2v}{1+v^2} dv = \int \frac{1}{x} dx$ $\tan^{-1}v - \frac{1}{2}\log(1+v^2) = \log|x| + C$ $\tan^{-1}(y/x) = \log|x| + \log\sqrt{1+y^2/x^2} + C = \log\sqrt{x^2+y^2} + C$ Final answer: $\tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2 + y^2} + C$

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