Related Formula
Exact differential: d((y)/(x)) = (x dy - y dx)/(x²)$$\text{Exact differential: } d\left(\frac{y}{x}\right) = \frac{x \, dy - y \, dx}{x^2}$$
Core Logic
Divide the entire differential equation by x²$x^2$ to construct an exact differential on the left hand side.
(x dy - y dx)/(x²) = √(x² + y²)x² dx$$\frac{x\, dy - y\, dx}{x^2} = \frac{\sqrt{x^2 + y^2}}{x^2} \, dx$$
Rewrite the right side by pushing an x$x$ under the square root:
d((y)/(x)) = √(1 + (y²)/(x²)) · (1)/(x) dx$$d\left(\frac{y}{x}\right) = \sqrt{1 + \frac{y^2}{x^2}} \cdot \frac{1}{x} \, dx$$
Step 1: Integration
Isolate the variables by dividing by the square root term:
∫ d((y)/(x))√(1 + ((y)/(x))²) = ∫ (1)/(x) dx$$\int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1 + \left(\frac{y}{x}\right)^2}} = \int \frac{1}{x} \, dx$$
Applying standard integral ∫ du√(1+u²) = ln(u + √(1+u²))$\int \frac{du}{\sqrt{1+u^2}} = \ln(u + \sqrt{1+u^2})$:
ln((y)/(x) + √(1 + (y²)/(x²))) = ln x + ln k = ln(kx)$$\ln\left(\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}\right) = \ln x + \ln k = \ln(kx)$$
Step 2: Finding Constant of Integration
Remove logarithms:
(y)/(x) + √(x² + y²)x = kx y + √(x² + y²) = kx²$$\frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = kx \implies y + \sqrt{x^2 + y^2} = kx^2$$
Apply the initial condition y(1) = 0$y(1) = 0$:
0 + √(1² + 0²) = k(1)² 1 = k$$0 + \sqrt{1^2 + 0^2} = k(1)^2 \implies 1 = k$$
The specific curve is y + √(x² + y²) = x²$y + \sqrt{x^2 + y^2} = x^2$.
Step 3: Calculating Final Value
Substitute x = 3$x = 3$ to find y(3)$y(3)$:
y + √(9 + y²) = 9$$y + \sqrt{9 + y^2} = 9$$
√(9 + y²) = 9 - y$$\sqrt{9 + y^2} = 9 - y$$
Square both sides:
9 + y² = 81 - 18y + y²$$9 + y^2 = 81 - 18y + y^2$$
18y = 72 y = 4$$18y = 72 \implies y = 4$$
Pattern Recognition
The expression x dy - y dx$x \, dy - y \, dx$ is the universal flag for the quotient rule exact differential d(y/x)$d(y/x)$. Dividing the whole equation by x²$x^2$ immediately separates variables.
Chapter Mix
Class 12 Maths: Differential Equations
