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For this separable differential equation, away from $x=1$, separation gives $\ln |y+1|=\ln |x-1|+C$, or $y+1=K(x-1)$ on a connected interval. The initial condition gives $2=-K$, henceThe maximal interval containing $x=0$ on which this solves the printed differential equation is $\boxed{-\infty <x<1}$. Although the linear formula extends through one, the right-hand side there is $0/0$ and the original equation is undefined. The equilibrium solution $y=-1$, lost if one divides by $y+1$, does not satisfy the initial condition.

This is a separable differential equation. Multiply by $e^{-y}$ and integrate:The initial condition gives $C=-1$. HenceFor a real solution the outer logarithm requires a positive argument. The maximal interval containing the initial point is therefore $\boxed{|x|<\sqrt{e^2-1}}$. At either endpoint $y$ diverges, so the real solution cannot continue through it with a finite value.

The integrating factor is $e^{kx}$, giving $(e^{kx}y{)}^{\prime }=a{e}^{kx}\sin (mx)$. The required antiderivative isAfter dividing by $e^{kx}$, the general solution is the oscillatory term plus $C{e}^{-kx}$. The initial condition fixes $C=1+am/(k^2+m^2)$, soThe real, nonzero parameters ensure $k^2+m^2>0$; direct substitution verifies both the equation and $y(0)=1$.

For a nonzero solution, the separable differential equation gives $dy/y=3x^{2}dx$. Integrating yields $\ln |y|=x^3+C$, hence $y=K{e}^{x^3}$. The initial condition fixes $K=3/e$, soThe zero solution omitted by division by $y$ cannot satisfy the given initial condition. Differentiating the displayed solution verifies $y^{\prime }=3x^{2}y$.

The characteristic equation is $r^2+r-2=(r-1)(r+2)=0$, giving $y=A{e}^x+B{e}^{-2x}$. The initial conditions give $A+B=1$ and $A-2B=-1$. Solving yields $A=1/3$, $B=2/3$, henceThe two distinct characteristic roots supply all solutions of this second-order linear ordinary differential equation.

The initial condition gives $e^{y(0)}=2=1+C$, so $C=1$. ConsequentlyIts logarithm has positive argument for every real $x$.

An integrating factor is $e^{3x}$. Multiplication gives $(e^{3x}y{)}^{\prime }=8e^{3x}$, so $y=8/3+C{e}^{-3x}$. The initial condition gives $C=4/3$. Therefore

The characteristic polynomial of the homogeneous equation is $r^2+7r+12=(r+3)(r+4)$. Thus the complementary function is $C_1{e}^{-3x}+C_2{e}^{-4x}$. The forcing resonates with the first term, so a resonant exponential particular solution is $y_p=Kx{e}^{-3x}$. Substitution gives $y_p^{\prime \prime }+7y_p^{\prime }+12y_p=K{e}^{-3x}$, hence $K=2$.

The full solution is $y=(C_1+2x)e^{-3x}+C_2{e}^{-4x}$. At zero, the two initial conditions give $C_1+C_2=1$ and $2-3C_1-4C_2=0$. Solving yields $C_1=2$, $C_2=-1$, so

The general solution of this first-order linear ordinary differential equation is $y=C{e}^{3x}$. Its initial condition gives $C=3$, soDifferentiation verifies $y^{\prime }=3y$ and substitution at zero gives the required value.

For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.

Impose the five initial conditions at zero on the preceding general solution. They giveThe second and fourth equations give $C_1=C_2$ and $C_4=0$. The third and fifth then yield $C_1=C_2=1/4$ and $C_3=-1/2$, while the first gives $C_0=1$. ThusFor a direct check, the difference of its fifth and first derivatives is $x$. At zero its value is one; its first, second, third and fourth derivatives vanish. In particular the quadratic term cancels the otherwise nonzero second derivative of the hyperbolic-cosine and cosine combination.

Separate variables near the nonzero initial condition: $y^{-2}dy=dx$. Integration gives $-1/y=x+C$, and $y(1)=1$ fixes $C=-2$. ThereforeThis is the maximal connected interval containing the initial point. Direct differentiation gives $y^{\prime }=1/(2-x{)}^2=y^2$; the pole at $x=2$ prevents continuation as a finite classical solution of a differential equation through that point.

Since $1+x^2>0$ for all real $x$, divide the equation by that factor. Its linear coefficient can be integrated by polynomial division:Thus a positive integrating factor is $M=e^{3x^2/2}(1+x^2{)}^{-3/2}$. Multiplying the divided equation gives the product rule formThe integration by substitution $u=1+x^2$ givesThe initial condition $y(0)=1$ fixes $C-2=1$. The required solution is thereforeNo real singularity was introduced by the division or the integrating factor, so the solution extends over the entire real line.

The separable differential equation gives $dy/y=-dx$, so $y=C{e}^{-x}$. The initial condition fixes $C=2$, and hence

The initial condition gives $1=1/(C-1)$, hence $C=2$. ThusThe displayed interval is the maximal connected interval containing the initial point $x=1$; the solution of a differential equation diverges at $x=2$.

This is a Bernoulli differential equation with exponent four. On the nonzero branch selected by $y(0)=1$, put $v=y^{-3}$. Multiplying the original equation by $-3y^{-4}$ givesIts integrating factor is $e^{-x}$, so $(e^{-x}v{)}^{\prime }=-3$ and $e^{-x}v=C-3x$. The initial condition sets $C=1$. ThusThis is the maximal real solution interval containing zero: the solution diverges as $x\to (1/3{)}^{-}$. The real negative branch of the same algebraic formula for $x>1/3$ is disconnected from this initial-value solution. The original equation also admits $y=0$, but it does not satisfy the prescribed initial 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.

To distinguish the one-variable profile from the resulting two-variable function, write $u(x,t)=F(\xi )$ with $\xi =x{e}^{-t}$. The chain rule gives $u_t=-x{e}^{-t}{F}^{\prime }(\xi )$ and $u_x=e^{-t}{F}^{\prime }(\xi )$, so $u_t+x{u}_x=0$. Thus any differentiable profile of $x{e}^{-t}$ solves the partial differential equation.

The initial condition selects $F(\xi )=\ln \xi$ on the positive half-line. HenceThis satisfies both the partial differential equation and the given data at $t=0$.

The integrating factor for this linear ordinary differential equation is $e^x$. Multiplication gives $(e^{x}y{)}^{\prime }=1$, so $e^{x}y=x+C$. The initial condition sets $C=0$, giving

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.