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Exercise 10.6 · Q8

Q.(x2+y2)dy=xy dx\left(x^2+y^2\right)dy=xy\,dx. It is given that y(1)=1y(1)=1 and y(x0)=ey(x_0)=e. Find the value of x0x_0.

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Concept understanding — Homogeneous Differential Equations

A function f(x,y)f(x,y) is a homogeneous function of degree nn if f(tx,ty)=tnf(x,y)f(tx,ty)=t^n f(x,y) for every suitably restricted x,y,tx,y,t (Euler's homogeneity). A homogeneous function of degree zero can always be written purely as a function of the single ratio yx\dfrac{y}{x} (or xy\dfrac{x}{y}): f(x,y)=g ⁣(yx)f(x,y)=g\!\left(\dfrac{y}{x}\right).

Homogeneous differential equation. An ODE is in homogeneous form if it can be written as

dydx=g ⁣(yx).\dfrac{dy}{dx}=g\!\left(\dfrac{y}{x}\right).

Equivalently, M(x,y) dx+N(x,y) dy=0M(x,y)\,dx+N(x,y)\,dy=0 is homogeneous exactly when MM and NN are homogeneous functions of the same degree — because then f(x,y)=−M/Nf(x,y)=-M/N is automatically homogeneous of degree 00. (This use of the word "homogeneous" for the equation is a different meaning from calling the constant term g(x)=0g(x)=0 in a linear equation "homogeneous" — Definition 10.7 versus Definition 10.12 in the textbook — so the two uses should not be confused.)

Solution method (Theorem 10.1). Substitute y=vxy=vx (so v=yxv=\dfrac{y}{x}), giving dydx=v+xdvdx\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}. The homogeneous equation becomes

v+xdvdx=g(v) ⟹ xdvdx=g(v)−v,v+x\dfrac{dv}{dx}=g(v)\ \Longrightarrow\ x\dfrac{dv}{dx}=g(v)-v,

which is variables-separable in vv and xx:

dvg(v)−v=dxx.\dfrac{dv}{g(v)-v}=\dfrac{dx}{x}.

Integrate both sides, then replace vv by yx\dfrac{y}{x} to return to the original variables. …

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