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For $x>0$ on a domain where the tangent is finite, put $v=y/x$. The homogeneous first-order differential equation becomes $x{v}^{\prime }=-\tan v$. Direct differentiation gives $(x\sin v{)}^{\prime }=0$, hence $x\sin (y/x)=C$, without dividing by $\sin v$ and losing constant-ratio solutions. If $C=x_0>0$ and $y/x\to \pi /2$ as $x\downarrow x_0$, two branches are $y=x\arcsin (x_0/x)$ and $y=x[\pi -\arcsin (x_0/x)]$, defined for $x>x_0$. They approach the same finite endpoint but have unbounded derivatives. The original tangent is undefined at the endpoint, so neither is a classical solution through that point. An implicit relation extending continuously to an initial point does not by itself solve an initial value problem there.

A homogeneous first-order differential equation has the form $y^{\prime }=F(y/x)$ on a domain where $x\ne 0$. The substitution $v=y/x$, so $y^{\prime }=v+x{v}^{\prime }$, reduces it toWhere $F(v)-v\ne 0$, this is a separable differential equation, with $\int dv/[F(v)-v]=\ln |x|+C$. Values $v_0$ satisfying $F(v_0)=v_0$ give the additional straight-line solutions $y=v_{0}x$, which must be checked before division. This scaling-based use of “homogeneous” is different from a homogeneous linear differential equation.

For example, $y^{\prime }=y/x+x/y$ gives $F(v)=v+1/v$, so $v{v}^{\prime }=1/x$ and $y^2=x^2(2\ln |x|+C)$. Real explicit branches are $y=\pm x\sqrt{2\ln |x|+C}$ on intervals where the expression under the square root is positive. Zeros of a denominator and endpoints of a branch must still be checked in the original ordinary differential equation.

The two affine lines meet at $(a,b)=(2,1)$. Translating their intersection to the origin, $X=x-2$, $Y=y-1$, removes both constant terms and gives the homogeneous first-order differential equationOn a patch with $X\ne 0$, put $Y=uX$. Then $Xdu/dX=(u^2+4)/(2-u)$, and separation of variables givesThus one local real implicit form isThe arctangent branch must be continued on the chosen solution patch. More geometrically, set $r=\sqrt{4X^2+Y^2}$ and take a continuous local argument $\theta =\arg (2X+iY)$; then $\theta -\ln r=C$. This form extends through $X=0$ where the original slope is defined. The intersection itself is excluded because the original right-hand side is $0/0$ there, and points with $2X-Y=0$ do not define a finite derivative. No real constant-$u$ solution was lost, because $u^2+4$ has no real zero. Translating the intersection works precisely because the two affine linear forms then become homogeneous linear forms.

On an interval with $x\ne 0$, this is a homogeneous first-order differential equation. Put $y=xv(x)$, so $y^{\prime }=v+x{v}^{\prime }$. Substitution and division by $x^2$ giveThe constant choices $v=0$ and $v=2$ give $y=0$ and $y=2x$. Otherwise the separable differential equation givesAbsorbing the sign and the exponential integration constant yields $(v-2)/v=C{x}^2$. Relabelling the arbitrary constant then givesThe first family already includes $y=2x$ at $K=0$. Choose any interval where its denominator is nonzero. These give all solutions on intervals avoiding $x=0$. Both families extend through $x=0$ with $y(0)=0$, as required by the original ordinary differential equation. Because it is singular there, differentiable continuations of the nonzero family may use different $K$ on the two sides: both one-sided derivatives at zero equal $2$.

For this homogeneous first-order differential equation, set $v=y/x$, with $x\ne 0$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution simplifies the equation toThe useful further substitution is $w=e^{-v}-v^2$. Its chain rule derivative satisfiesThus $(x^{2}w{)}^{\prime }=0$. Replacing $v$ and $w$ gives the implicit solutionNo division by $w$ was needed, so the level $C=0$ is included. A direct verification also identifies the integrating factor for a differential one-form:Setting this exact differential to zero recovers precisely the original equation after multiplication by the nonzero factor $e^{y/x}$. The coefficient $1+2v{e}^v$ never vanishes for real $v$, since $v{e}^v\ge -1/e$ implies $1+2v{e}^v\ge 1-2/e>0$. Hence these level sets give local function graphs $y(x)$ everywhere with $x\ne 0$.

On an interval avoiding zero put $v=y/x$. A real solution requires $|v|\le 1$, and the principal square root is $\sqrt{x^2-y^2}=|x|\sqrt{1-v^2}$. The homogeneous first-order differential equation reduces toWhere $|v|<1$, separation of variables gives $\arcsin v=\mathrm{sgn}(x)\log |x|+C$. Thus the nonconstant pieces arerestricted to intervals where the cosine of the phase is nonnegative. This sign condition is essential when substituting back into the original square root.

There are also the solutions $y=x$ and $y=-x$. They can join the nonconstant pieces with a continuous first derivative, because the transformed rate vanishes but is not Lipschitz continuous at $v=\pm 1$. For example, on the positive axis every $A>0$ givesThe sticking solutions of a bounded square-root differential equation describe the same joining mechanism on either half-axis after using the increasing variable $t=\mathrm{sgn}(x)\log |x|$.

The homogeneous first-order differential equation suggests $v=y/x$, so $y^{\prime }=v+x{v}^{\prime }$. Wherever the original equation is defined,Thus the implicit solution is $x\sin (y/x)=C$. The printed initial values force $C=x_0=2/\pi$.

There is no classical solution through the printed initial point. There $y/x=\pi /2$, so the tangent on the right is undefined. The implicit relation nevertheless has two neighboring real branches on $x>x_0$ which approach that point continuously:Both tend to $y=1$ as $x\downarrow x_0$, but their derivatives become unbounded with opposite signs. They solve the equation on the open interval $x>x_0$, not at its endpoint. Other inverse-sine periods cannot approach the specified initial value. This is an endpoint obstruction in a homogeneous differential equation. Constant-ratio branches $y=k\pi x$ also solve the general equation where $x\ne 0$, but do not meet the initial data.

Work on an interval with $x\ne 0$. The homogeneous first-order differential equation becomes a separable differential equation under $v=y/x$: substituting $y=xv$ gives $v+x{v}^{\prime }=v^2+v$, hence $x{v}^{\prime }=v^2$.

For $v\ne 0$, integration gives $-1/v=\ln |x|+C$. Renaming the arbitrary constant producesFor the nonzero family the denominator must remain nonzero, so take any connected interval avoiding both $x=0$ and $|x|=e^C$. The zero solution must be restored after division by $v^2$; by the Picard-Lindelöf theorem away from $x=0$, a solution cannot switch between zero and a nonzero branch there.

For real solutions with $x>0$, the square root requires $y\ge 0$. First retain the solution $y\equiv 0$, which would be lost by dividing by a square root of $y$.

For $y>0$, use the homogeneous first-order differential equation substitution $v=\sqrt{y/x}>0$, so $y=x{v}^2$, $\sqrt{xy}=xv$ and $y^{\prime }=v^2+2xv{v}^{\prime }$. The equation becomesThis is now a separable differential equation. Integration gives $-2/v+2\ln v=-\ln x+C$. Since $x{v}^2=y$, the positive solutions satisfyThe implicit family is sufficient here because only parts (a) and (b) request an explicit expression. It also shows why the nonnegative choice of $v$ matters: $\sqrt{xy}$ denotes the positive square root.

Because the equation depends on the ratio $y/x$, set $u=y/x$, so $y^{\prime }=u+x{u}^{\prime }$. Dividing by $x^2$ and substituting givesThis reduces the homogeneous first-order differential equation to a separable differential equation. For $u\ne 1$,At $x=e$, the condition $y=0$ gives $u=0$, so $1=1+C$ and $C=0$. HenceThe interval displayed is the maximal solution interval containing $x=e$: there is a pole at $x=1$. Separating variables omitted the solution $u=1$, or $y=x$, but that solution does not satisfy the given condition.

The equation is not separable, because $x^2-xy=x(x-y)$ cannot be written as a product of a function of $x$ and a function of $y$ on an open region. For example, the ratio $(x^2-x{y}_1)/(x^2-x{y}_2)=(x-y_1)/(x-y_2)$ depends on $x$ for distinct fixed $y_1,y_2$, whereas a separated product would give a constant ratio.

It is not homogeneous in the sense of a homogeneous first-order differential equation: the right-hand side is not a function of $y/x$ alone. It is also not a homogeneous linear differential equation, since moving the $xy$ term to the left leaves a nonzero forcing.

It is linear, since it has the standard linear ordinary differential equation formThus the classifications are (a) no, (b) no, (c) yes.

Use the substitution $v=y/x$ for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, it givesWhere $\sin v\ne 0$, this separable differential equation integrates to $\ln |\sin v|=\ln |x|+C$, hence $\sin v=C_{1}x$. One can obtain the same relation without division by $\sin v$: the equation gives $(\sin v{)}^{\prime }=(\sin v)/x$, so $\sin v=C_{1}x$, including $\sin v\equiv 0$.

All real branches of the inverse sine are captured byEach branch is used on an interval with $x\ne 0$ and $|Cx|<1$, with fixed $k$. The strict inequality excludes zeros of $\cos (y/x)$, where the original tangent is undefined. When $C=0$, the formula includes all straight-line solutions $y=k\pi x$. Keeping both families of inverse sine branches is essential for the two different initial conditions below.

For $x\ne 0$ and $y\ne 0$, put $v=y/x$. This is the standard substitution for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, the equation becomesTaking antiderivatives yields $v^2=2\ln |x|+C$. The explicit general solution is thereforeon any interval where $x\ne 0$ and $2\ln |x|+C>0$, with a fixed choice of sign. For a check that does not divide by $y$, set $w=y^2$ to get $w^{\prime }-2w/x=2x$ and hence $w=x^2(2\ln |x|+C)$.

There is no missing zero solution: at $y=0$ with $x\ne 0$, the original equation would require $x^2=0$. Also no real solution on an interval about $x=0$ can arise from this family, since $2\ln |x|+C$ becomes negative near $0$.

An implicit homogeneous first-order differential equation $y/x=F(p)$, where $p=dy/dx$, can be reduced by treating the slope as an auxiliary variable. Differentiating $y=xF(p)$ givesWhere the divisions are valid, this is a separable differential equation. Constant slopes lost when dividing by $p-F(p)$ must be checked in the original equation.

For $F(p)=2p/(1-p^2)$, the reduced equation is $2x{p}^{\prime }=-p(1-p^2)$ and its nonconstant solutions obey $x{p}^2/(1-p^2)=C$. Substitution recovers the parabolas $y^2=4C(x+C)$, together with the constant solution $y=0$. Real graph branches must have $p^2=C/(x+C)\ge 0$, and the original restrictions $x\ne 0$, $p\ne \pm 1$ still apply. The constant slopes $p=\pm 1$ solve the reduced equation but are excluded from the original equation.