Differential Equations: Diagram-Based Questions (9)

hard • 3 min read
Tags Differential Equations

Question

A curve passes through the point (1,1)(1, 1) such that the slope of its tangent at any point (x,y)(x, y) equals y(x+y)x(y−x)\tfrac{y(x+y)}{x(y-x)}. Find the equation of the curve.

Solution — Step by Step

dydx=y(x+y)x(y−x)=xy+y2xy−x2\tfrac{dy}{dx} = \tfrac{y(x+y)}{x(y-x)} = \tfrac{xy + y^2}{xy - x^2}

Both numerator and denominator are homogeneous of degree 22, so this is a homogeneous DE. Substitute y=vxy = vx.

dydx=v+xdvdx\tfrac{dy}{dx} = v + x\tfrac{dv}{dx}. The RHS becomes:

x(vx)+(vx)2x(vx)−x2=vx2+v2x2vx2−x2=v+v2v−1\tfrac{x(vx) + (vx)^2}{x(vx) - x^2} = \tfrac{vx^2 + v^2x^2}{vx^2 - x^2} = \tfrac{v + v^2}{v - 1} v+xdvdx=v+v2v−1v + x\tfrac{dv}{dx} = \tfrac{v + v^2}{v - 1} xdvdx=v+v2v−1−v=v+v2−v(v−1)v−1=v+v2−v2+vv−1=2vv−1x\tfrac{dv}{dx} = \tfrac{v + v^2}{v - 1} - v = \tfrac{v + v^2 - v(v-1)}{v - 1} = \tfrac{v + v^2 - v^2 + v}{v - 1} = \tfrac{2v}{v - 1} (v−1) dv2v=dxx\tfrac{(v - 1)\,dv}{2v} = \tfrac{dx}{x}

LHS: 12∫(1−1v)dv=12(v−ln⁡∣v∣)\tfrac{1}{2}\int \left(1 - \tfrac{1}{v}\right)dv = \tfrac{1}{2}(v - \ln|v|).

RHS: ln⁡∣x∣+C\ln|x| + C.

12(v−ln⁡∣v∣)=ln⁡∣x∣+C\tfrac{1}{2}(v - \ln|v|) = \ln|x| + C v−ln⁡∣v∣=2ln⁡∣x∣+2Cv - \ln|v| = 2\ln|x| + 2C

At x=1,y=1x = 1, y = 1: v=1v = 1, so 1−ln⁡1=2ln⁡1+2C1 - \ln 1 = 2 \ln 1 + 2C, giving C=1/2C = 1/2.

Substitute back v=y/xv = y/x:

yx−ln⁡∣yx∣=2ln⁡∣x∣+1\tfrac{y}{x} - \ln\left|\tfrac{y}{x}\right| = 2\ln|x| + 1 yx=ln⁡∣y∣−ln⁡∣x∣+2ln⁡∣x∣+1=ln⁡∣xy∣+1\tfrac{y}{x} = \ln|y| - \ln|x| + 2\ln|x| + 1 = \ln|xy| + 1

Final answer: yx=ln⁡∣xy∣+1\mathbf{\tfrac{y}{x} = \ln|xy| + 1}, or equivalently y=x(1+ln⁡∣xy∣)y = x(1 + \ln|xy|).

Why This Works

Homogeneous DEs are recognized by checking that both numerator and denominator have the same degree when treated as polynomials in xx and yy. The substitution y=vxy = vx converts them to a separable form in vv and xx.

Why does this work? Homogeneity means f(tx,ty)=tkf(x,y)f(tx, ty) = t^k f(x, y). Setting t=1/xt = 1/x replaces every yy by y/x=vy/x = v and the xx factors out, leaving an expression purely in vv.

Alternative Method

Could try substituting x=vyx = vy instead — gives a different but equivalent equation. Useful if the algebra in the first substitution gets ugly. Whichever variable’s coefficient is cleaner in the denominator, that’s usually the better substitution.

Common Mistake

After substituting y=vxy = vx, students often forget the vv on the LHS (from dy/dx=v+xdv/dxdy/dx = v + x dv/dx). Missing this turns a separable equation into nonsense.

Homogeneous DEs are a steady 44-mark CBSE board question and 11 MCQ in JEE Main. The pattern is always the same: recognize, substitute y=vxy = vx, separate, integrate, back-substitute, apply IC. Practise the algebra so it’s automatic.

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