Chapter 9 · Differential Equations

9.3 Homogeneous Differential Equations

When the slope $\tfrac{dy}{dx}$ depends only on the ratio $\tfrac yx$, the equation is homogeneous: its slope field looks the same at every magnification. The substitution $y=vx$ turns it into an equation with variables separable.

Try to separate the variables in $\dfrac{dy}{dx}=\dfrac{x^2+3y^2}{2xy}$. You cannot: the numerator is a sum, and no algebra splits the right-hand side into (function of $x$) × (function of $y$). But look at it differently. Divide the top and bottom by $x^2$: $$\dfrac{dy}{dx}=\dfrac{1+3(y/x)^2}{2(y/x)}.$$ The slope depends on $x$ and $y$ only through the ratio $\dfrac yx$ — that is, only through the direction of the point $(x,y)$ as seen from the origin. Every point on the line $y=2x$ has the same slope, $\tfrac{13}{4}$, whether it is $(1,2)$ or $(1000,2000)$. If you photograph the slope field and then zoom in or out about the origin, the picture does not change at all.

Such equations are called homogeneous, and the scale symmetry suggests the cure: make the ratio itself the unknown. Writing $y=vx$, where $v=\dfrac yx$ is a new unknown function of $x$, turns every homogeneous equation into one with variables separable, which Section 9.2 taught you to solve. Homogeneous equations describe shapes that have no built-in scale — the curve of a car headlight's mirror, the paths of a pursuit, flows around a corner — and they appear in every Board paper as a 4- or 5-mark question.

Homogeneous Functions and Homogeneous Equations

📘 Definition 9.7 — Homogeneous function of degree $n$

A function $F(x,y)$ is a homogeneous function of degree $n$ if $$F(\lambda x,\lambda y)=\lambda^nF(x,y)$$ for every non-zero constant $\lambda$ (and all $(x,y)$ where both sides are defined). Equivalently, $F(x,y)=x^n\,g\!\left(\dfrac yx\right)$ for some function $g$ (or $F(x,y)=y^n\,h\!\left(\dfrac xy\right)$).

For example $x^2-3xy+5y^2$ is homogeneous of degree $2$ (every term has total degree $2$), and $\dfrac{x-y}{x+y}$ and $\sin\dfrac yx$ are homogeneous of degree $0$.

📘 Definition 9.8 — Homogeneous differential equation

A differential equation $\dfrac{dy}{dx}=F(x,y)$ is called a homogeneous differential equation if $F(x,y)$ is a homogeneous function of degree zero; then $F(x,y)=g\!\left(\dfrac yx\right)$ and the equation reads $$\dfrac{dy}{dx}=g\!\left(\dfrac yx\right).$$ Likewise $\dfrac{dx}{dy}=G(x,y)$ is homogeneous if $G$ is homogeneous of degree zero, $G(x,y)=h\!\left(\dfrac xy\right)$.

In particular, $\dfrac{dy}{dx}=\dfrac{P(x,y)}{Q(x,y)}$ is homogeneous whenever $P$ and $Q$ are homogeneous functions of the same degree.

✏️ Example 1 — Testing for homogeneity

Decide whether each function is homogeneous, and if so give its degree: (a) $x^2+3xy-y^2$ (b) $\dfrac{x^3+y^3}{x+y}$ (c) $x^2+y$ (d) $\sin\dfrac yx+\dfrac{x}{y}$. Hence decide whether $\dfrac{dy}{dx}=\dfrac{x^2+3xy-y^2}{x^2}$ and $\dfrac{dy}{dx}=\dfrac{x^2+y}{x}$ are homogeneous equations.

💡 Show solution▶

(a) $F(\lambda x,\lambda y)=\lambda^2x^2+3\lambda^2xy-\lambda^2y^2=\lambda^2F(x,y)$: homogeneous of degree $2$.

(b) $\dfrac{\lambda^3(x^3+y^3)}{\lambda(x+y)}=\lambda^2\cdot\dfrac{x^3+y^3}{x+y}$: degree $2$ (indeed it simplifies to $x^2-xy+y^2$).

(c) $(\lambda x)^2+\lambda y=\lambda^2x^2+\lambda y$, which is not $\lambda^n(x^2+y)$ for any single $n$ (try $x=0$ and $x=y=1$): not homogeneous. The terms have different degrees.

(d) $\sin\dfrac{\lambda y}{\lambda x}+\dfrac{\lambda x}{\lambda y}=F(x,y)=\lambda^0F(x,y)$: homogeneous of degree $0$.

Equations The first right-hand side is (degree $2$)/(degree $2$), of degree $0$: a homogeneous equation, $\dfrac{dy}{dx}=1+3\dfrac yx-\left(\dfrac yx\right)^2$. The second is not, since $x^2+y$ is not homogeneous.

The grey dashes are the slope field. Play sweeps a ray from the origin around the plane; the dashes at points on the ray are drawn in red. For a homogeneous equation they are all parallel — the slope depends only on $v=\tfrac yx$ — and the zoom slider changes nothing. Switch to the non-homogeneous equation $\dfrac{dy}{dx}=x+y$ and watch both properties fail.

Equation:zoom:

ray direction $v=\tfrac yx$—

smallest slope on the ray—

largest slope on the ray—

window half-width—

Solving by the Substitution y = vx

📐 Theorem 9.5 — The substitution $y=vx$

Let $g$ be continuous on an interval, and consider the homogeneous equation $\dfrac{dy}{dx}=g\!\left(\dfrac yx\right)$ on an interval of $x$ not containing $0$. Put $y=vx$, where $v$ is a new unknown function of $x$. Then $$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx},$$ and the equation becomes $$x\dfrac{dv}{dx}=g(v)-v,$$ which has variables separable. Where $g(v)\ne v$: $$\int\dfrac{dv}{g(v)-v}=\int\dfrac{dx}{x}+C=\log|x|+C.$$ After integrating, replace $v$ by $\dfrac yx$. Similarly, $\dfrac{dx}{dy}=h\!\left(\dfrac xy\right)$ is solved by $x=vy$, which gives $y\dfrac{dv}{dy}=h(v)-v$.

Proof. If $y=vx$, the product rule gives $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$, and $\dfrac yx=v$. Substituting both into $\dfrac{dy}{dx}=g\!\left(\dfrac yx\right)$ gives $v+x\dfrac{dv}{dx}=g(v)$, i.e. $\dfrac{dv}{dx}=\dfrac1x\cdot\big(g(v)-v\big)$ — a product of a function of $x$ and a function of $v$. Theorem 9.2 finishes the job. $\blacksquare$

A number $c$ with $g(c)=c$ gives the straight-line solution $y=cx$ (then $v$ is constant): look for these lost solutions, as in Section 9.2.

🧭 Problem-solving strategy — homogeneous equations

Write the equation as $\dfrac{dy}{dx}=F(x,y)$ and check that $F(\lambda x,\lambda y)=F(x,y)$ (degree $0$). Board answers expect this check to be shown.
Put $y=vx$ and $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. Simplify the right-hand side: every $x$ should cancel, leaving a function of $v$.
Separate: $\dfrac{dv}{g(v)-v}=\dfrac{dx}{x}$, and integrate (partial fractions or a standard integral may be needed).
Replace $v$ by $\dfrac yx$, simplify, and apply any initial condition.
If the equation is more naturally $\dfrac{dx}{dy}=h\!\left(\dfrac xy\right)$ — for instance when $e^{x/y}$ appears — use $x=vy$ instead.

✏️ Example 2 — A first substitution

Solve $x\dfrac{dy}{dx}=y-x$ for $x\gt0$, and find the solution with $y(1)=2$.

💡 Show solution▶

Check $\dfrac{dy}{dx}=\dfrac{y-x}{x}$, and $\dfrac{\lambda y-\lambda x}{\lambda x}=\dfrac{y-x}{x}$: homogeneous, with $g(v)=v-1$.

Substitute $y=vx$: $v+x\dfrac{dv}{dx}=v-1$, so $x\dfrac{dv}{dx}=-1$ and $dv=-\dfrac{dx}{x}$.

Integrate $v=-\log x+C$, so $\dfrac yx=C-\log x$: $$y=Cx-x\log x.$$

Particular $y(1)=C=2$: $y=2x-x\log x$.

Check $y'=2-\log x-1=1-\log x$, and $\dfrac{y-x}{x}=2-\log x-1=1-\log x$ ✓.

✏️ Example 3 — A family through the origin

Find the general solution of $2xy\dfrac{dy}{dx}=x^2+3y^2$ ($x\gt0$, $y\gt0$).

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Check $\dfrac{dy}{dx}=\dfrac{x^2+3y^2}{2xy}$: numerator and denominator are both homogeneous of degree $2$, so the equation is homogeneous, with $g(v)=\dfrac{1+3v^2}{2v}$.

Substitute $x\dfrac{dv}{dx}=\dfrac{1+3v^2}{2v}-v=\dfrac{1+v^2}{2v}$.

Separate and integrate $\dfrac{2v\,dv}{1+v^2}=\dfrac{dx}{x}$, so $\log(1+v^2)=\log x+C$, i.e. $1+v^2=Ax$ with $A=e^C\gt0$.

Back to $y$ $1+\dfrac{y^2}{x^2}=Ax$, so $$x^2+y^2=Ax^3.$$

Check Differentiating, $2x+2yy'=3Ax^2=\dfrac{3(x^2+y^2)}{x}$, so $2xyy'=3x^2+3y^2-2x^2=x^2+3y^2$ ✓.

✏️ Example 4 — A trigonometric homogeneous equation

Solve $\dfrac{dy}{dx}=\dfrac yx+\tan\dfrac yx$ ($x\gt0$) with $y(1)=\dfrac\pi6$.

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Substitute $g(v)=v+\tan v$, so $x\dfrac{dv}{dx}=\tan v$ and $\cot v\,dv=\dfrac{dx}{x}$.

Integrate $\log|\sin v|=\log x+C$, so $\sin v=Ax$, i.e. $\sin\dfrac yx=Ax$.

Particular At $x=1$: $\sin\dfrac\pi6=A$, so $A=\tfrac12$ and $\sin\dfrac yx=\dfrac x2$.

Explicit With $\dfrac yx$ starting at $\tfrac\pi6$ inside $\left(-\tfrac\pi2,\tfrac\pi2\right)$: $y=x\sin^{-1}\dfrac x2$, valid for $0\lt x\lt2$ (at $x=2$, $\tfrac yx=\tfrac\pi2$ and $\tan\tfrac yx$ is undefined).

Check At $x=1.2$: $y'=\sin^{-1}0.6+\dfrac{0.6}{\sqrt{0.64}}$ $\approx0.6435+0.75=1.3935$, and $\dfrac yx+\tan\dfrac yx=0.6435+\tan0.6435$ $\approx0.6435+0.75$ ✓.

✏️ Example 5 — An exponential homogeneous equation

Solve $x\dfrac{dy}{dx}=y+xe^{y/x}$ ($x\gt0$) with $y(1)=0$.

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Substitute $\dfrac{dy}{dx}=\dfrac yx+e^{y/x}$, so $g(v)=v+e^v$ and $x\dfrac{dv}{dx}=e^v$.

Separate and integrate $e^{-v}\,dv=\dfrac{dx}{x}$: $-e^{-v}=\log x+C$.

Particular At $x=1$, $v=0$: $-1=C$. So $e^{-y/x}=1-\log x$.

Explicit Taking logs, $y=-x\log(1-\log x)$, valid while $1-\log x\gt0$, i.e. $0\lt x\lt e$. (As $x\to e^-$, $y\to+\infty$.)

✏️ Example 6 — When x = vy is the better substitution

Solve $\dfrac{dx}{dy}=\dfrac xy+\dfrac yx$ ($x\gt0$, $y\gt0$), given that $x=\sqrt2$ when $y=1$.

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Recognise The equation gives $\dfrac{dx}{dy}$ as $h\!\left(\dfrac xy\right)$ with $h(v)=v+\dfrac1v$. Put $x=vy$, so $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$.

Separate $v+y\dfrac{dv}{dy}=v+\dfrac1v$, so $v\,dv=\dfrac{dy}{y}$.

Integrate $\dfrac{v^2}{2}=\log y+C$, i.e. $x^2=2y^2(\log y+C)$.

Particular $2=2(0+C)$, so $C=1$: $x^2=2y^2(1+\log y)$.

Check $2x\dfrac{dx}{dy}=4y(1+\log y)+2y=\dfrac{2x^2}{y}+2y$, so $\dfrac{dx}{dy}=\dfrac xy+\dfrac yx$ ✓.

Pick an example and press Next step (or Play) to walk through the substitution. Set the starting value $y(1)=y_0$ with the slider: the red curve is the formula obtained from the steps, and the blue dots are an independent numerical (Runge–Kutta) solution of the original equation from the same starting point. If the algebra is right, they coincide.

Example:$y_0$:

Homogeneous: $\dfrac{dy}{dx}=\dfrac yx-1=g(v)$ with $g(v)=v-1$.
Put $y=vx$: $\ v+x\dfrac{dv}{dx}=v-1$.
Separate: $\ dv=-\dfrac{dx}{x}$.
Integrate: $\ v=C-\log x$.
Back-substitute: $\ y=Cx-x\log x$; $\ y(1)=y_0$ gives $C=y_0$.

step—

start $(1,y_0)$ and constant—

formula at $x_1$—

numerical at $x_1$—

difference—

⚠️ Common mistakes when studying alone

1. Writing $\dfrac{dy}{dx}=\dfrac{dv}{dx}$. With $y=vx$, both factors vary: $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ (product rule). Forgetting the $v$ is the most common error in this topic.

2. Not checking homogeneity. $\dfrac{dy}{dx}=\dfrac{x+y+1}{x-y}$ is not homogeneous (the $1$ has degree $0$, the other terms degree $1$); $y=vx$ will not remove $x$. If $x$ does not cancel after substituting, the equation was not homogeneous — or you made an algebra slip.

3. Forgetting to substitute back. The answer must be in terms of $x$ and $y$, not $v$.

4. Mishandling the constant. $\log(1+v^2)=\log x+C$ gives $1+v^2=Ax$, not $1+v^2=x+C$.

5. Using $y=vx$ when $x=vy$ is easier. If the equation contains $\tfrac xy$ (for example $e^{x/y}$) or is given as $\tfrac{dx}{dy}=\dots$, put $x=vy$ and $\tfrac{dx}{dy}=v+y\tfrac{dv}{dy}$.

6. Ignoring the interval. Solutions such as $y=x\sin^{-1}\tfrac x2$ or $y=-x\log(1-\log x)$ exist only on a limited interval; state it when asked for "the solution".

Exercises

Homogeneous or not? Give the degree of each homogeneous function.

1 $x^2-2xy+5y^2$
2 $\dfrac{x^3-y^3}{x^2+y^2}$
3 $x^2+y^2+1$
4 $\dfrac yx+\cos\dfrac yx$
5 $x\sin\dfrac yx-y$
6 Which of these differential equations are homogeneous? (a) $\dfrac{dy}{dx}=\dfrac{x+2y}{2x-y}$ (b) $\dfrac{dy}{dx}=\dfrac{x+y+1}{x-y}$ (c) $\dfrac{dy}{dx}=\left(\dfrac yx\right)^2+e^{y/x}$

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1 Every term has degree $2$: homogeneous of degree $2$.

2 $\dfrac{\lambda^3(x^3-y^3)}{\lambda^2(x^2+y^2)}$: homogeneous of degree $1$.

3 Not homogeneous: $\lambda^2(x^2+y^2)+1\ne\lambda^n(x^2+y^2+1)$ in general.

4 A function of $\tfrac yx$ only: homogeneous of degree $0$.

5 $\lambda x\sin\dfrac yx-\lambda y=\lambda F(x,y)$: degree $1$.

6 (a) and (c). In (b) the constant $1$ spoils homogeneity.

Find the general solution ($x\gt0$ where a logarithm of $x$ appears).

1 $\dfrac{dy}{dx}=\dfrac{x+2y}{x}$
2 $2xy\dfrac{dy}{dx}=x^2+y^2$
3 $\dfrac{dy}{dx}=\dfrac yx+\dfrac{y^2}{x^2}$
4 $\dfrac{dy}{dx}=\dfrac{y-x}{y+x}$
5 $\dfrac{dy}{dx}=\dfrac yx+\sec\dfrac yx$
6 $(x^2+y^2)\dfrac{dy}{dx}=xy$
7 $2x^2\dfrac{dy}{dx}=(x+y)^2$
8 $\dfrac{dx}{dy}=\dfrac{x+y}{y}$ ($y\gt0$)
9 $xy^2\dfrac{dy}{dx}=x^3+y^3$

💡 Show solutions▶

1 $x\dfrac{dv}{dx}=1+v$: $\log|1+v|=\log x+C$, $1+v=Ax$, so $x+y=Ax^2$, i.e. $y=Ax^2-x$.

2 $g(v)=\dfrac{1+v^2}{2v}$, $x\dfrac{dv}{dx}=\dfrac{1-v^2}{2v}$: $\dfrac{2v\,dv}{1-v^2}=\dfrac{dx}{x}$ gives $-\log|1-v^2|=\log x+C$, so $(1-v^2)x=A$, i.e. $x^2-y^2=Ax$.

3 $x\dfrac{dv}{dx}=v^2$: $-\dfrac1v=\log x+C$, so $y=-\dfrac{x}{\log x+C}$ (and the lost solution $y=0$).

4 $g(v)=\dfrac{v-1}{v+1}$, $g(v)-v=-\dfrac{1+v^2}{1+v}$: $\dfrac{(1+v)\,dv}{1+v^2}=-\dfrac{dx}{x}$ gives $\tan^{-1}v+\tfrac12\log(1+v^2)=-\log|x|+C$, i.e. $\tan^{-1}\dfrac yx+\tfrac12\log(x^2+y^2)=C$.

5 $x\dfrac{dv}{dx}=\sec v$: $\cos v\,dv=\dfrac{dx}{x}$, so $\sin\dfrac yx=\log x+C$.

6 $g(v)=\dfrac{v}{1+v^2}$, $g(v)-v=-\dfrac{v^3}{1+v^2}$: $\dfrac{1+v^2}{v^3}dv=-\dfrac{dx}{x}$ gives $-\dfrac{1}{2v^2}+\log|v|=-\log|x|+C$, i.e. $\log|y|-\dfrac{x^2}{2y^2}=C$.

7 $g(v)=\tfrac12(1+v)^2$, $g(v)-v=\tfrac12(1+v^2)$: $\dfrac{2\,dv}{1+v^2}=\dfrac{dx}{x}$, so $2\tan^{-1}\dfrac yx=\log x+C$.

8 $x=vy$: $v+y\dfrac{dv}{dy}=v+1$, so $v=\log y+C$ and $x=y\log y+Cy$.

9 $g(v)=\dfrac{1+v^3}{v^2}$, $g(v)-v=\dfrac{1}{v^2}$: $v^2\,dv=\dfrac{dx}{x}$, so $\dfrac{v^3}{3}=\log x+C$, i.e. $y^3=3x^3(\log x+C)$.

III

Particular solutions ($x\gt0$, $y\gt0$ unless stated).

1 $x\dfrac{dy}{dx}=2y-x$, $y(1)=3$
2 $x^2\dfrac{dy}{dx}=y^2+2xy$, $y(1)=1$
3 $\dfrac{dy}{dx}=\dfrac yx-\dfrac xy$, $y(1)=2$
4 $x\dfrac{dy}{dx}=y+x\cos^2\dfrac yx$, $y(1)=\dfrac\pi4$
5 $y\dfrac{dx}{dy}=x+ye^{-x/y}$, $x=0$ when $y=1$

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1 $x\dfrac{dv}{dx}=v-1$: $v-1=Ax$, so $y=x+Ax^2$; $3=1+A$, $A=2$: $y=x+2x^2$.

2 $x\dfrac{dv}{dx}=v^2+v$: $\dfrac{dv}{v(v+1)}=\dfrac{dx}{x}$ gives $\log\dfrac{v}{v+1}=\log x+C$, so $\dfrac{y}{x+y}=Ax$. At $(1,1)$, $A=\tfrac12$: $2y=x(x+y)$, i.e. $y=\dfrac{x^2}{2-x}$ ($0\lt x\lt2$).

3 $x\dfrac{dv}{dx}=-\dfrac1v$: $\dfrac{v^2}{2}=-\log x+C$; $C=2$. So $y^2=x^2(4-2\log x)$, valid for $0\lt x\lt e^2$.

4 $x\dfrac{dv}{dx}=\cos^2v$: $\sec^2v\,dv=\dfrac{dx}{x}$, $\tan v=\log x+C$; $\tan\tfrac\pi4=1=C$. So $\tan\dfrac yx=1+\log x$.

5 $x=vy$: $y\dfrac{dv}{dy}=e^{-v}$, $e^v\,dv=\dfrac{dy}{y}$, $e^{x/y}=\log y+C$; $C=1$. So $e^{x/y}=1+\log y$.

Curves, slopes and reasoning.

1 Find the curve through $(2,1)$ whose slope at every point $(x,y)$ is $\dfrac{x^2+y^2}{2xy}$.
2 Find the curve through $(1,2)$ whose slope at every point $(x,y)$ ($x\gt0$) is $\dfrac{2y}{x}+\dfrac xy$.
3 Explain why the substitution $y=vx$ does not help with $\dfrac{dy}{dx}=\dfrac{x+y+1}{x-y}$.
4 For $\dfrac{dy}{dx}=\dfrac{y-x}{y+x}$, find the lines through the origin along which the slope is $0$, along which it is $-1$, and along which the tangent is vertical.
5 Prove that if $y=\varphi(x)$ is a solution of $\dfrac{dy}{dx}=g\!\left(\dfrac yx\right)$, then so is $y=\lambda\varphi\!\left(\dfrac x\lambda\right)$ for every constant $\lambda\gt0$. (This is the zoom symmetry of Interactive 1.)

💡 Show solutions▶

1 By II.2, $x^2-y^2=Ax$. At $(2,1)$: $3=2A$, $A=\tfrac32$. So $2x^2-2y^2=3x$.

2 $x\dfrac{dv}{dx}=v+\dfrac1v=\dfrac{v^2+1}{v}$: $\dfrac{v\,dv}{1+v^2}=\dfrac{dx}{x}$, so $1+v^2=Ax^2$, i.e. $x^2+y^2=Ax^4$. At $(1,2)$, $A=5$: $x^2+y^2=5x^4$.

3 The right-hand side is not homogeneous of degree $0$: with $y=vx$ it becomes $\dfrac{x(1+v)+1}{x(1-v)}$, which still contains $x$, so the variables do not separate.

4 With $v=\tfrac yx$ the slope is $\dfrac{v-1}{v+1}$. It is $0$ when $v=1$ (the line $y=x$); $-1$ when $v-1=-v-1$, $v=0$ (the $x$-axis); vertical when $v=-1$ (the line $y=-x$).

5 Let $u(x)=\lambda\varphi(x/\lambda)$. Then $u'(x)=\varphi'(x/\lambda)$ $=g\!\left(\dfrac{\varphi(x/\lambda)}{x/\lambda}\right)$ $=g\!\left(\dfrac{\lambda\varphi(x/\lambda)}{x}\right)$ $=g\!\left(\dfrac{u(x)}{x}\right)$ ✓. Enlarging a solution curve by the factor $\lambda$ about the origin gives another solution curve.

🎯 Exam corner Board · 1 markBoard · 3 marksBoard · 5 marksJEE MainJEE Advanced

How this is examined: Board papers almost always contain one homogeneous equation for 3 or 5 marks — show it is homogeneous, substitute $y=vx$ (or $x=vy$), separate, integrate and apply the initial condition. Marks are lost for skipping the homogeneity check and for not substituting back. JEE asks for a value of the solution or a property of the solution curve.

E1. Board · 1 mark Which of the following is a homogeneous differential equation? (A) $\dfrac{dy}{dx}=\dfrac{x+y+1}{x-y}$ (B) $\dfrac{dy}{dx}=\dfrac{x^2+y}{xy}$ (C) $\dfrac{dy}{dx}=\dfrac{x^2+y^2}{xy}$ (D) $\dfrac{dy}{dx}=x+y^2$

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Idea Numerator and denominator must be homogeneous of the same degree. In (C) both have degree $2$. In (A) and (B) the numerator mixes degrees; (D) mixes degrees $1$ and $2$.

Answer (C).

E2. Board · 1 mark The most suitable substitution for solving $\dfrac{dx}{dy}=\dfrac xy+e^{x/y}$ is (A) $y=vx$ (B) $x=vy$ (C) $x+y=v$ (D) $xy=v$

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Idea The equation gives $\dfrac{dx}{dy}$ as a function of $\dfrac xy$, so put $x=vy$, $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$; it becomes $y\dfrac{dv}{dy}=e^v$.

Answer (B). (A) also works in principle but leads to a much messier equation.

E3. Board · 3 marks Show that $x\dfrac{dy}{dx}=y-x\tan\dfrac yx$ is homogeneous and find its general solution ($x\gt0$).

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Homogeneous $\dfrac{dy}{dx}=F(x,y)=\dfrac yx-\tan\dfrac yx$ and $F(\lambda x,\lambda y)=F(x,y)$: degree $0$. (1 mark)

Substitute $y=vx$: $v+x\dfrac{dv}{dx}=v-\tan v$, so $\cot v\,dv=-\dfrac{dx}{x}$. (1 mark)

Answer $\log|\sin v|=-\log x+C$, so $x\sin\dfrac yx=A$. (1 mark)

E4. Board · 5 marks Find the particular solution of $x^2\dfrac{dy}{dx}=x^2+xy+y^2$ ($x\gt0$) given that $y=0$ when $x=1$.

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Homogeneous $\dfrac{dy}{dx}=1+\dfrac yx+\left(\dfrac yx\right)^2$, a function of $\dfrac yx$ (degree $0$). (1 mark)

Substitute $y=vx$: $v+x\dfrac{dv}{dx}=1+v+v^2$, so $x\dfrac{dv}{dx}=1+v^2$. (1 mark)

Separate and integrate $\dfrac{dv}{1+v^2}=\dfrac{dx}{x}$: $\tan^{-1}v=\log x+C$. (1 mark)

Constant At $x=1$, $v=0$: $C=0$. So $\tan^{-1}\dfrac yx=\log x$. (1 mark)

Answer $y=x\tan(\log x)$, valid for $e^{-\pi/2}\lt x\lt e^{\pi/2}$ (where $\log x$ stays in the principal range of $\tan^{-1}$). (1 mark)

E5. JEE Main The solution curve of $\dfrac{dy}{dx}=\dfrac yx+\dfrac xy$ ($x\gt0$, $y\gt0$) through $(1,1)$ meets the line $x=e$ at $y=$ (A) $\sqrt3\,e$ (B) $\sqrt2\,e$ (C) $3e$ (D) $e$

💡 Show solution▶

Method $y=vx$: $x\dfrac{dv}{dx}=\dfrac1v$, so $v\,dv=\dfrac{dx}{x}$ and $\dfrac{v^2}{2}=\log x+C$. At $(1,1)$: $C=\tfrac12$. So $y^2=x^2(1+2\log x)$.

Answer At $x=e$: $y^2=3e^2$, $y=\sqrt3\,e$: (A). Trap (B) forgets the factor $2$ in $\tfrac{v^2}{2}$.

E6. JEE Advanced (One or more correct options.) Let $y(x)$ satisfy $x\dfrac{dy}{dx}=y+\sqrt{x^2-y^2}$ for $x\ge1$, with $y(1)=0$, on the interval where $\tfrac yx$ increases from $0$ to $1$. Then (A) $y=x\sin(\log x)$ on that interval (B) $y\!\left(e^{\pi/4}\right)=\dfrac{e^{\pi/4}}{\sqrt2}$ (C) the curve passes through $\left(e^{\pi/6},\tfrac12e^{\pi/6}\right)$ (D) $y=x\sin(\log x)$ satisfies the equation for every $x\gt1$

💡 Show solution▶

Solve $y=vx$: $x\dfrac{dv}{dx}=\sqrt{1-v^2}$, so $\dfrac{dv}{\sqrt{1-v^2}}=\dfrac{dx}{x}$, $\sin^{-1}v=\log x+C$, and $C=0$. Hence $v=\sin(\log x)$ for $1\le x\le e^{\pi/2}$: (A) ✓.

Values (B) $e^{\pi/4}\sin\tfrac\pi4=\dfrac{e^{\pi/4}}{\sqrt2}$ ✓. (C) $e^{\pi/6}\sin\tfrac\pi6=\tfrac12e^{\pi/6}$ ✓.

(D) For $y=x\sin(\log x)$ we get $xy'-y=x\cos(\log x)$, but $\sqrt{x^2-y^2}=x|\cos(\log x)|$. These differ when $\cos(\log x)\lt0$, e.g. $x=e^{2}$. So (D) is false.

Answer (A), (B), (C). The trap is (D): $\sqrt{1-\sin^2\theta}=|\cos\theta|$, not $\cos\theta$.

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. What shape must a mirror have so that every ray of light from a bulb at the origin is reflected parallel to the $x$-axis — the design of a torch, a car headlight or a satellite dish (run in reverse)?

Why it matters. Headlights, solar cookers, radio telescopes and the dish antennas used for DTH television all rely on this shape. The reflection law gives a homogeneous differential equation, because the problem has no built-in scale: enlarging a working mirror about the bulb gives another working mirror.

What to do.

Let a ray from $O$ hit the mirror at $P(x,y)$, making angle $\theta$ with the $x$-axis, so $\tan\theta=\tfrac yx$; let the tangent at $P$ make angle $\alpha$ with the $x$-axis, so $\tan\alpha=\tfrac{dy}{dx}=m$. Draw a diagram and use "angle of incidence = angle of reflection" to explain why $\theta=2\alpha$ when the reflected ray is horizontal.
Use $\tan2\alpha=\dfrac{2m}{1-m^2}$ to show that $\dfrac yx=\dfrac{2m}{1-m^2}$, and solve this quadratic for $m$: $\dfrac{dy}{dx}=\dfrac{-x+\sqrt{x^2+y^2}}{y}$ (taking the root that gives a mirror behind the bulb). Check that the right-hand side is homogeneous of degree $0$.
Verify, by differentiating, that every parabola $y^2=2Cx+C^2$ ($C\gt0$) satisfies this equation. Where is its vertex, and where is the bulb relative to it?
Measure a real reflector (a torch, or a photo of a dish antenna with a known diameter): its width $D$ across the rim and depth $d$. For a parabola the bulb (focus) should sit at distance $f=\dfrac{D^2}{16d}$ from the vertex. Check this against the actual bulb position.

Tools: Ruler, protractor, a torch or a photo of a dish antenna, graph paper or a graphing app.

Hand in: Your diagram and derivation of the differential equation, the verification for $y^2=2Cx+C^2$, a plot of three members of the family, and your measurement with the predicted and actual focus positions.

Going further. Solve the equation yourself with the substitution $x=vy$, or by writing $\sqrt{x^2+y^2}=r$ and noticing that $x\,dx+y\,dy=r\,dr$. Why do all members of the family have the same shape (just enlarged)?

Instructor note: the verification is $2yy'=2C$, so $y'=\tfrac Cy$, and $\tfrac{-x+\sqrt{x^2+y^2}}{y}=\tfrac{-x+(x+C)}{y}=\tfrac Cy$ because $x^2+y^2=(x+C)^2$ on the parabola. The vertex is $\left(-\tfrac C2,0\right)$, so the focal length is $\tfrac C2$. A dish $60$ cm across and $10$ cm deep has $f=\tfrac{3600}{160}=22.5$ cm.

🧪 Self-check quiz

0/5 correct

Q1. The degree of the homogeneous function $x^3+2x^2y-y^3$ is

Q2. Which function is homogeneous of degree $0$?

Q3. If $y=vx$, where $v$ is a function of $x$, then $\dfrac{dy}{dx}=$

Q4. After the substitution $y=vx$, the equation $\dfrac{dy}{dx}=1+\dfrac yx$ becomes

Q5. For a homogeneous equation $\dfrac{dy}{dx}=g\!\left(\dfrac yx\right)$, the slope is the same at all points of

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