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For $y\ne 0$, the separable differential equation gives $dy/y^2=dx$. Integrating gives $-1/y=x+C_0$, orEach solution is considered on an interval not containing its pole. The additional stationary solution $y=0$ is excluded by the stated condition.

For the solution through $x=0$, work on the connected domain $x>-1$. While $y>0$, separation of variables givesThe initial condition gives $C=2$, soThe restriction matters: the unsquared equation requires $1-\frac{1}{2}\ln (1+x)\ge 0$. The solution reaches zero at $x_{\ast }=e^2-1$. Continuing the squared formula beyond that point would give a positive derivative, contradicting the original negative square-root right-hand side.

The nonnegative forward continuation is insteadThe two pieces join with $y^{\prime }=0$, and both satisfy the original equation. Because the right-hand side is nonpositive for $x>-1$, a nonnegative solution cannot leave zero in the forward direction. This is a finite-time extinction phenomenon for a separable differential equation.

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.

Near the initial condition $y=1$, divide by $y^2$ and separate variables: $y^{-2}dy=-x^{2}dx$. Integration gives $-1/y=-x^3/3+C$; the initial value fixes $C=-1$. Thus the solution of the separable differential equation isThe stated interval is the maximal connected interval containing zero: the solution diverges at its left endpoint. The constant solution $y=0$ does not satisfy the initial condition.

For the Gaussian ansatz, partial differentiation givesCancel the positive exponential and compare the constant and quadratic powers of $x$ in the diffusion equation. The two coupled ordinary differential equations areThe initial profile requires $F(0)=T_0$ and $H(0)=L^{-2}$. The second equation is a separable differential equation:Substitute this into the first equation and integrate its logarithmic derivative:Thusand the solution isThis is Gaussian diffusion of a smooth initial profile: its width grows while its peak decreases. The Gaussian integral gives ${\int }_{-\infty }^{\infty }T(x,t)dx=\sqrt{\pi }{T}_{0}L$, independent of time, which checks conservation of the total amount under the diffusion equation.

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.

On an interval with $x\ne 0$, division by $x^2$ gives the separable differential equation $y^{\prime }=(4/x^2-1/x)y^2$. For $y\ne 0$, take antiderivatives:Renaming the constant yields the explicit nonzero familyFor $x\ne 0$, the interval of definition must avoid zeros of the denominator. Division by $y^2$ excluded the additional solution $\boxed{y(x)\equiv 0}$, which must also be included.

The original undivided differential equation also makes sense at $x=0$, where it requires $y(0)=0$. For continuation through a zero of the leading ODE coefficient, rewrite the nonzero family asIt extends differentiably with $y(0)=0$ and $y^{\prime }(0)=1/4$, since $y(x)/x\to 1/4$. One may choose the integration constants independently on the two sides of zero and obtain a differentiable solution through zero. The zero solution also extends through zero, with derivative $0$; joining a nonzero branch on just one side to a zero branch on the other would fail differentiability.

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.

On an interval with $x\ne 0$, the equation is a separable differential equation. For $y\ne 0$,Taking antiderivatives gives $-1/y=-4/x-x^3/3+C$. Renaming the constant, the explicit general nonzero solution isIt is defined on any interval avoiding $x=0$ and zeros of its denominator. The separation divided by $y^2$, so the additional solution $\boxed{y(x)\equiv 0}$ must be included.

Where $\cos x\ne 0$, the separable differential equation is $y^{\prime }=y\tan x$. Separating and taking antiderivatives gives $\ln |y|=-\ln |\cos x|+C$, for nonzero $y$. Including the zero solution, the explicit general solution ison any interval with $\cos x\ne 0$. It can also be obtained without dividing by $y$: the product rule gives $(y\cos x{)}^{\prime }=0$.

A nonzero solution cannot extend through a zero of $\cos x$. If a differentiable solution is required on an interval containing such a zero, the original equation there requires $y=0$, and the only such solution across that zero is $y\equiv 0$.

The product rule gives $v^{\prime }=x{y}^{\prime }+y$ for $v=xy$, reducing the displayed equation to the autonomous separable differential equation $v^{\prime }=F(v)$. On an interval where $F(v)\ne 0$, integrate $\int dv/F(v)=x+C$ and recover $y=v/x$ for $x\ne 0$. Zeros of $F$ give additional constant solutions for $v$ and must be considered before division. A solution for $y$ extends through zero only if the recovered quotient has the necessary finite derivatives and satisfies the undivided original equation there.

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.