What is homogeneous differential equation?

DE is homogeneous if dy/dx = f(y/x). Example: dy/dx = (y+√(x²+y²))/x can be written as function of y/x. Solve using v=y/x substitution.

How to solve homogeneous DE?

Use v = y/x, so y = vx and dy/dx = v + x(dv/dx). Substitute in original equation, separate variables, integrate, apply initial condition.

What is sinh⁻¹ integration?

∫(1/√(1+v²))dv = sinh⁻¹(v) + C = ln(v + √(1+v²)) + C. Inverse hyperbolic sine function, commonly appears in homogeneous DE solutions.

How to calculate sinh(ln x)?

sinh(ln x) = (e^(ln x) - e^(-ln x))/2 = (x - 1/x)/2 = (x² - 1)/(2x). This simplification gives final solution y = (x² - 1)/2.

Solution curve xdy - ydx = √(x² + y²)dx, y(1)=0, find y(3)

Solution curve xdy – ydx = √(x² + y²)dx, y(1)=0, find y(3) | JEE Main 2026

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JEE 2026 Math

Q707 MCQ Differential Eq

Let the solution curve of differential equation $xdy – ydx = \sqrt{x^2 + y^2}\,dx$, $x > 0$, $y(1) = 0$ be $y = y(x)$. Then $y(3)$ equals

✅ Correct Answer

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Solution Steps

Rewrite equation in standard form

$xdy – ydx = \sqrt{x^2 + y^2}\,dx$

$xdy = ydx + \sqrt{x^2 + y^2}\,dx$

$\dfrac{dy}{dx} = \dfrac{y + \sqrt{x^2 + y^2}}{x}$

Check for homogeneous form

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

This depends only on $\dfrac{y}{x}$ → Homogeneous DE

Apply substitution v = y/x

Let $v = \dfrac{y}{x}$, so $y = vx$

$\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

Substituting: $v + x\dfrac{dv}{dx} = v + \sqrt{1 + v^2}$

$x\dfrac{dv}{dx} = \sqrt{1 + v^2}$

Separate variables and integrate

$\dfrac{dv}{\sqrt{1 + v^2}} = \dfrac{dx}{x}$

$\displaystyle\int \dfrac{dv}{\sqrt{1 + v^2}} = \int \dfrac{dx}{x}$

$\sinh^{-1}(v) = \ln x + C$

Use initial condition y(1) = 0

When $x = 1$, $y = 0$ → $v = 0$

$\sinh^{-1}(0) = \ln(1) + C$

$0 = 0 + C$ → $C = 0$

So: $\sinh^{-1}(v) = \ln x$

Solve for v

$v = \sinh(\ln x)$

$v = \dfrac{e^{\ln x} – e^{-\ln x}}{2} = \dfrac{x – \frac{1}{x}}{2} = \dfrac{x^2 – 1}{2x}$

Get y and find y(3)

$y = vx = x \cdot \dfrac{x^2 – 1}{2x} = \dfrac{x^2 – 1}{2}$

$y(3) = \dfrac{3^2 – 1}{2} = \dfrac{8}{2} = \boxed{4}$

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Final Answer

y(3) = 4

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Related Theory

📌 Homogeneous Differential Equations

A first-order differential equation is homogeneous if it can be written in form dy/dx = f(y/x), where right side depends only on ratio y/x. Alternatively, equation Mdx + Ndy = 0 is homogeneous if both M(x,y) and N(x,y) are homogeneous functions of same degree. A function f(x,y) is homogeneous of degree n if f(tx,ty) = t^n·f(x,y) for all t. To solve: use substitution v = y/x, which means y = vx. Differentiating gives dy/dx = v + x(dv/dx). Substitute into original equation to get separable form in v and x. After separation, integrate both sides: ∫g(v)dv = ∫h(x)dx. Finally substitute v = y/x back to get solution. Common examples include dy/dx = (y/x) + tan(y/x), or x²(dy/dx) = y² + xy. Recognition is key: look for expressions that can be written in terms of y/x ratio. In our problem, dividing numerator and denominator by x revealed dependence on y/x only.

📌 Separable Differential Equations

A DE is separable if it can be written as dy/dx = g(x)·h(y), allowing separation: (1/h(y))dy = g(x)dx. Integration of both sides gives ∫(1/h(y))dy = ∫g(x)dx + C. The key is separating variables to opposite sides. Examples: dy/dx = xy separates to (dy/y) = x dx. After integration: ln|y| = x²/2 + C, giving y = Ae^(x²/2). For dy/dx = e^(x-y), rewrite as e^y dy = e^x dx, then integrate. Homogeneous equations become separable after v = y/x substitution. In our case, x(dv/dx) = √(1+v²) separates to dv/√(1+v²) = dx/x. This transforms complex homogeneous DE into manageable separable form. Separation method is fundamental as it reduces DE to two independent integrations. However, not all DEs are separable – some require integrating factors or other techniques.

📌 Hyperbolic Functions & Integration

Hyperbolic functions are analogs of trig functions: sinh(x) = (e^x – e^(-x))/2, cosh(x) = (e^x + e^(-x))/2, tanh(x) = sinh(x)/cosh(x). Identity: cosh²x – sinh²x = 1. Derivatives: d/dx(sinh x) = cosh x, d/dx(cosh x) = sinh x. Inverse functions: sinh⁻¹(x) = ln(x + √(x²+1)), cosh⁻¹(x) = ln(x + √(x²-1)) for x≥1. Key integral: ∫dx/√(1+x²) = sinh⁻¹(x) + C = ln(x + √(1+x²)) + C. Similarly ∫dx/√(x²-1) = cosh⁻¹(x) + C. In our problem, ∫dv/√(1+v²) gives sinh⁻¹(v). To evaluate sinh(ln x), use definition: sinh(ln x) = (e^(ln x) – e^(-ln x))/2 = (x – 1/x)/2 = (x²-1)/(2x). This appears frequently in homogeneous DEs. Hyperbolic functions model catenary curves, special relativity, and appear in many physics applications. Understanding their properties and integrals is crucial for JEE.

📌 Initial Value Problems (IVP)

An IVP consists of differential equation plus initial condition y(x₀) = y₀. General solution of first-order DE contains one arbitrary constant C. Initial condition determines C uniquely, giving particular solution. Process: (1) Solve DE to get general solution with C, (2) Apply y(x₀) = y₀ to find C, (3) Write particular solution. For example, dy/dx = 2x with y(0) = 1 gives general solution y = x² + C. Using y(0) = 1: 1 = 0 + C, so C = 1 and particular solution is y = x² + 1. Existence-Uniqueness Theorem: if f(x,y) and ∂f/∂y are continuous near (x₀,y₀), then unique solution exists. In our problem, y(1) = 0 gave v = 0 at x = 1, leading to C = 0. This determined specific curve among infinite family of solutions. IVPs model real scenarios: initial temperature, starting position, launch velocity. They’re fundamental in physics, engineering, and JEE problems.

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FAQs

How to recognize homogeneous DE?

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Check if dy/dx = f(y/x). Divide all terms by appropriate power of x to see if it reduces to function of y/x only. If yes → homogeneous. Use v = y/x substitution.

Why does v = y/x work?

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Since equation depends on y/x ratio, replacing this ratio with single variable v simplifies structure. Gives dy/dx = v + x(dv/dx), converting to separable form.

What is sinh⁻¹ and its integral?

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Inverse hyperbolic sine. ∫(1/√(1+v²))dv = sinh⁻¹(v) + C = ln(v + √(1+v²)) + C. Both forms equivalent, use whichever is convenient.

How to simplify sinh(ln x)?

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sinh(ln x) = (e^(ln x) – e^(-ln x))/2 = (x – 1/x)/2 = (x² – 1)/(2x). Key step to get final answer y = (x² – 1)/2.

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