NCERT Solutions for Class 12 Maths Differential Equations Exercise 9.5

Understanding Homogeneous Differential Equations

A differential equation of the form $\frac{dy}{dx} = F(x, y)$ is said to be homogeneous if $F(x, y)$ is a homogeneous function of degree zero. Mathematically, a function $f(x, y)$ is homogeneous of degree $n$ if $f(\lambda x, \lambda y) = \lambda^n f(x, y)$ for any non-zero constant $\lambda$. For Exercise 9.5, we focus exclusively on equations where $n=0$, implying $F(\lambda x, \lambda y) = F(x, y)$. This special property guarantees that the equation can be simplified by expressing it as a function of the single ratio $\frac{y}{x}$ or $\frac{x}{y}$. By using the change of variable substitution $y = vx$ (which transforms the derivative to $\frac{dy}{dx} = v + x\frac{dv}{dx}$), we convert a complicated coupled differential equation into a simple Variable Separable form. Once separated, simple integration yields the general or particular solution.

Step-by-Step Algorithm to Solve Homogeneous Differential Equations

1. Express and Verify Homogeneity — Write the given equation in the standard form: $\frac{dy}{dx} = F(x, y)$. Replace $x$ with $\lambda x$ and $y$ with $\lambda y$. If $F(\lambda x, \lambda y) = F(x, y)$, the equation is homogeneous.
2. Substitute the Variable — Substitute $y = vx$. Differentiate this relation with respect to $x$ using the product rule to get: $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
3. Separate the Variables — Substitute the expressions for $y$ and $\frac{dy}{dx}$ into the original equation. Simplify the algebraic terms to isolate the variables $v$ and $x$ on opposite sides of the equation.
4. Integrate and Substitute Back — Integrate both sides of the separated equation. Finally, replace $v$ with $\frac{y}{x}$ to obtain the general solution in terms of $x$ and $y$.

CBSE Board Exam Tips & Common Traps

Keep these expert tips in mind to avoid losing marks:

• The Alternative Substitution: If the differential equation contains terms of the form $\frac{x}{y}$ or is written as $\frac{dx}{dy} = F(x, y)$, use the substitution $x = vy$ instead of $y = vx$. The corresponding derivative is $\frac{dx}{dy} = v + y\frac{dv}{dy}$.
• Integrating Constant: Always add the integration constant $C$ immediately after integrating. Leaving it out or adding it late is a very common mistake.
• Logarithmic Simplifications: When terms integrate to natural logarithms, it is often useful to write the constant of integration as $\ln C$ instead of $C$. This allows you to apply logarithm rules like $\ln A + \ln B = \ln(AB)$ to simplify your final answer into a neat algebraic form.

Practice Questions with Solutions

• Q: Show that the differential equation $(x^2 + xy)dy = (x^2 + y^2)dx$ is homogeneous and solve it. A: Step 1: Write the equation as $\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}$. Let $F(x, y) = \frac{x^2 + y^2}{x^2 + xy}$. Finding $F(\lambda x, \lambda y) = \frac{\lambda^2 x^2 + \lambda^2 y^2}{\lambda^2 x^2 + \lambda^2 xy} = \frac{x^2 + y^2}{x^2 + xy} = F(x, y)$. Since the degree of $\lambda$ is $0$, the equation is homogeneous. Step 2: Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Substituting these in the equation: $v + x\frac{dv}{dx} = \frac{x^2 + v^2 x^2}{x^2 + v x^2} = \frac{1 + v^2}{1 + v}$. Step 3: Separate variables: $x\frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v = \frac{1 + v^2 - v - v^2}{1 + v} = \frac{1 - v}{1 + v}$. Rearranging gives: $\frac{1 + v}{1 - v} dv = \frac{dx}{x}$. Step 4: Integrate both sides: $\int \frac{1+v}{1-v} dv = \int \frac{dx}{x}$. Using substitution or splitting the numerator: $\int (\frac{2}{1-v} - 1) dv = \int \frac{dx}{x}$. $-2\ln|1-v| - v = \ln|x| + C$. Substitute $v = \frac{y}{x}$: $-2\ln|1 - \frac{y}{x}| - \frac{y}{x} = \ln|x| + C$. Final answer: $\ln|x| - 2\ln|\frac{x-y}{x}| - \frac{y}{x} = C$, which simplifies to $y + x\ln\frac{(x-y)^2}{x} = C'x$.
• Q: Solve the homogeneous differential equation: $(x-y)dy - (x+y)dx = 0$. A: Step 1: Rewrite as $\frac{dy}{dx} = \frac{x+y}{x-y}$. Let $F(x,y) = \frac{x+y}{x-y}$. We can check $F(\lambda x, \lambda y) = \frac{\lambda(x+y)}{\lambda(x-y)} = F(x,y)$. Thus, it is a homogeneous differential equation. Step 2: Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$: $v + x\frac{dv}{dx} = \frac{x+vx}{x-vx} = \frac{1+v}{1-v}$. Step 3: Isolate variables: $x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v - v + v^2}{1-v} = \frac{1+v^2}{1-v}$. Separating variables: $\frac{1-v}{1+v^2} dv = \frac{dx}{x}$. Step 4: Integrate both sides: $\int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv = \int \frac{dx}{x}$. $\tan^{-1}(v) - \frac{1}{2}\ln|1+v^2| = \ln|x| + C$. Substitute $v = \frac{y}{x}$: $\tan^{-1}(\frac{y}{x}) - \frac{1}{2}\ln|1 + \frac{y^2}{x^2}| = \ln|x| + C$. $\tan^{-1}(\frac{y}{x}) - \frac{1}{2}\ln|\frac{x^2+y^2}{x^2}| = \ln|x| + C$. $\tan^{-1}(\frac{y}{x}) - \frac{1}{2}\ln|x^2+y^2| + \ln|x| = \ln|x| + C$. Final answer: $\tan^{-1}(\frac{y}{x}) - \frac{1}{2}\ln(x^2+y^2) = C$.
• Q: Solve the differential equation: $x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy$. A: Step 1: Rewrite as $\frac{dy}{dx} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x}$. This shows the right side is a function of $\frac{y}{x}$, hence it is homogeneous. Step 2: Put $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$: $v + x\frac{dv}{dx} = 1 - 2v^2 + v$. Step 3: Simplify and separate variables: $x\frac{dv}{dx} = 1 - 2v^2$. $\frac{dv}{1-2v^2} = \frac{dx}{x}$. Step 4: Integrate both sides: $\int \frac{dv}{1 - (\sqrt{2}v)^2} = \int \frac{dx}{x}$. Using the formula $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \ln|\frac{a+x}{a-x}|$, we get: $\frac{1}{2\sqrt{2}} \ln|\frac{1 + \sqrt{2}v}{1 - \sqrt{2}v}| = \ln|x| + C$. Substitute $v = \frac{y}{x}$: Final answer: $\frac{1}{2\sqrt{2}} \ln|\frac{x + \sqrt{2}y}{x - \sqrt{2}y}| = \ln|x| + C$.
• Q: Find the particular solution of the differential equation $x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0$, given that $y = \pi$ when $x = 1$. A: Step 1: Rewrite as $x \frac{dy}{dx} = y - x \sin\left(\frac{y}{x}\right) \implies \frac{dy}{dx} = \frac{y}{x} - \sin\left(\frac{y}{x}\right)$. Since it is in the form of a function of $\frac{y}{x}$, it is homogeneous. Step 2: Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$: $v + x\frac{dv}{dx} = v - \sin(v) \implies x\frac{dv}{dx} = -\sin(v)$. Step 3: Separate variables and integrate: $\csc(v) dv = -\frac{dx}{x}$. $\int \csc(v) dv = -\int \frac{dx}{x}$. $\ln|\csc(v) - \cot(v)| = -\ln|x| + \ln|C| \implies \ln|\csc(v) - \cot(v)| = \ln|\frac{C}{x}|$. $\csc(v) - \cot(v) = \frac{C}{x}$. Substitute $v = \frac{y}{x}$: $\csc\left(\frac{y}{x}\right) - \cot\left(\frac{y}{x} ight) = \frac{C}{x}$. Step 4: Apply boundary conditions ($y = \pi$, $x = 1$): $\csc(\pi) - \cot(\pi)$ is undefined, let's simplify first: $\frac{1 - \cos(v)}{\sin(v)} = \tan(v/2) = \frac{C}{x}$. Substituting $y = \pi, x = 1 \implies v = \pi/1 = \pi$. $\tan(\pi/2)$ is undefined. Let's re-evaluate using the original integration form: $\ln|\tan(v/2)| = -\ln|x| + \ln|C| \implies x\tan(\frac{y}{2x}) = C$. For $y = \pi, x = 1 \implies 1 \cdot \tan(\pi/2)$ is undefined. Let us write the alternative form: $\cot(\frac{y}{2x}) = C x$. When $y = \pi, x = 1 \implies \cot(\pi/2) = 0 \implies C = 0$. Final answer: The particular solution is $\cos(\frac{y}{2x}) = 0$, or simply $y = \pi x$.

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