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The characteristic equation of the associated homogeneous linear ordinary differential equation is $m^2+5m+6=(m+2)(m+3)=0$, so the complementary function is $C_1{e}^{-2x}+C_2{e}^{-3x}$. For a particular solution, try $y_p=ax+b$. Substitution gives $6ax+5a+6b=6x$, so $a=1$, $b=-5/6$. The unrestricted solution is thereforeIts derivative at zero is $-2C_1-3C_2+1$. The single prescribed condition imposes $2C_1+3C_2=1$ and leaves one arbitrary constant. Thus the most general permitted solution isThere is no condition fixing $y(0)$, so the remaining arbitrary constant is essential.

Write $y(\theta )=Y(x)$ with $x=\cos \theta$. By the chain rule,The terms involving $Y_x$ cancel in the original ordinary differential equation, leavingOn each interval where $\sin \theta \ne 0$, the transformed linear ordinary differential equation is $Y_{xx}+2Y=0$. Its characteristic equation has roots $\pm i\sqrt{2}$, yieldingDirect substitution shows these expressions also extend smoothly through $\theta =k\pi$ and satisfy the original equation there. The two constants describe the general solution on any interval where the coordinate substitution is invertible.

On the interval $x>-1$ containing the initial condition, division by $1+x$ gives a linear ordinary differential equation with integrating factor $(1+x{)}^{-3}$. ConsequentlyThe initial value gives $C=1$. ThusThis polynomial also satisfies the original equation at $x=-1$ and gives an extension to all real $x$. The initial value problem is uniquely determined on $x>-1$; the vanishing leading coefficient at $-1$ does not ensure unique continuation to its left.

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.

For the first-order linear ordinary differential equation, an integrating factor isMultiplication converts the left side to a derivative:Integrating gives $(1+x^2{)}^{2}y=\arctan x+C$. The initial value fixes $C=1$, so

First write the forcing as $\sin x\cos x$. An integrating factor for the linear ordinary differential equation is $e^{-\sin x}$, givingSet $u=\sin x$ inside the antiderivative. ThenThe initial value forces $C=1$, henceThe resulting expression is explicit and valid for all real $x$; the substitution does not impose a global monotonicity condition on $\sin x$ when verifying an antiderivative.

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.

An integrating factor isMultiplying the linear ordinary differential equation by it givesThe right-hand side integrates to $(a^2+x^2{)}^{3/2}/3$, soThis holds on every interval where the coefficient is defined. If $a=0$, intervals cannot cross $x=0$; equivalently the family is $y=x^2/3+C/x$ on either half-line, with an independently chosen constant.

For the first-order linear ordinary differential equation, seek an integrating factor $M(x)$ that makes its left side a product rule derivative. Multiplying by $M$ gives $M{y}^{\prime }+MPy$, and comparison with $(My{)}^{\prime }=M{y}^{\prime }+M^{\prime }y$ requires $M^{\prime }=PM$. On an interval where $P$ is continuous chooseThis factor is never zero. The transformed equation is $(My{)}^{\prime }=MQ$, so integration from $x_0$ to $x$ yieldsHere $C=y(x_0)$ for an initial value problem. The indefinite integral version is $y=e^{-\int Pdx}[C+\int Q{e}^{\int Pdx}dx]$, with any constant choices inside the integrating factor absorbed into $C$. Continuity of $P,Q$ is a convenient sufficient hypothesis for a classical differentiable solution of a differential equation.

The original differential equation is defined only where $x\ne 0$ and $y\ne 0$. Put $w=y^3$, so the chain rule gives $w^{\prime }=3y^2{y}^{\prime }$. Multiplying the equation by $3y^2$ yields a first-order linear ordinary differential equation:An integrating factor on either half-line is $1/x$, up to an irrelevant nonzero constant. Equivalently, apply the quotient rule directly:Integrating gives $w/x=x^2+C$. Thus the real general solution ison any connected interval where $x\ne 0$ and $x^3+Cx\ne 0$. These restrictions preserve both the original denominator and differentiability of the real cube root. In particular $C=0$ includes $y=x$ on either half-line; this solution has not been lost through division by an expression involving $y/x-1$.

For verification, differentiating $y^3=x^3+Cx$ gives $3y^2{y}^{\prime }=3x^2+C$, while $(2x^3+y^3)/x=3x^2+C$. The two sides agree wherever the original equation is defined.

For $x\ne 0$, divide by $x^2$ to obtain $y^{\prime }+y/x=1/x^2$, a first-order linear ordinary differential equation. An integrating factor is $x$ on either interval not crossing zero, since the product rule givesIntegration and division by $x$ give the explicit general solutionThe original equation would read $0=1$ at zero, so no solution can include that point.

Rewrite the linear ordinary differential equation as $y^{\prime }+y=e^{3x}$. Multiplication by its integrating factor $e^x$ gives $(e^{x}y{)}^{\prime }=e^{4x}$. Integrating yieldsThe general solution consists of this particular solution and the homogeneous solution $C{e}^{-x}$.

Under $y=xu$, the product rule gives $y^{\prime }=u+x{u}^{\prime }$ and $y^{\prime \prime }=2u^{\prime }+x{u}^{\prime \prime }$. Substitute these into the original equation:Since $x>0$, this is precisely $x{u}^{\prime \prime }+u^{\prime }=0$, or $(x{u}^{\prime }{)}^{\prime }=0$. Integrating twice gives $x{u}^{\prime }=C_1$ and $u=C_1\ln x+C_2$. ThereforeThe two independent solutions of a differential equation supply the general solution of this second-order linear ordinary differential equation.

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

The real equation has domain $x>0$. Its integrating factor isThus the linear ordinary differential equation becomes $(x^{x}y{)}^{\prime }=x^x{x}^{-x}=1$. After integration, $x^{x}y=x+C$. The initial condition gives $2=1+C$, hence

For $x\ne 0$, multiply by $x$ to obtain the first-order linear ordinary differential equation $y^{\prime }+xy=5x{e}^{-x^2}$. Its integrating factor is $\mu =e^{x^2/2}$. The product rule then givesIntegration yields $y{e}^{x^2/2}=-5e^{-x^2/2}+C$, soThe constant of integration is arbitrary on each connected interval of the original equation's domain. The formula extends smoothly through zero, but the printed equation itself contains $1/x$ and is not defined there.

For a first-order linear ordinary differential equation $y^{\prime }+p(x)y=q(x)$, an integrating factor is $\exp (\int p(x)dx)$. Here $p(x)=\cot x$, soThe sine is positive on the specified interval, so this integrating factor never vanishes there. Any nonzero constant multiple is equally suitable.

The homogeneous linear differential equation is $y^{\prime \prime }+y=0$. Its characteristic equation is ${\lambda }^2+1=0$, with roots $\pm i$. The general real complementary function is thereforeThe two trigonometric solutions are linearly independent, supplying the two arbitrary constants required for this second-order linear ordinary differential equation.

First retain the equilibrium solution $\boxed{y\equiv 0}$. Any nonzero real solution has constant sign on its interval of existence, since the polynomial right-hand side is locally Lipschitz and Picard-Lindelöf theorem prevents crossing the zero solution.

For a nonzero solution, the Bernoulli differential equation substitution $u=y^{-4}$ givesThis linear ordinary differential equation has solution $u+2=C{e}^{-8x}$. Therefore every nonzero real solution isEither sign may be chosen and remains fixed. The restriction ensures $y^{-4}>0$; nonpositive $C$ gives no nonzero real solution. These are maximal real intervals, and their restrictions are also solutions. As the right endpoint is approached, $|y|$ diverges in finite-time blowup. The substitution would have lost $y=0$ if it had not been checked separately.

On a nonconstant branch with $\sin y\ne 0$, regard $x$ as a function of $y$. The equation becomes the linear ordinary differential equationAn integrating factor is $\sin y$, givingTaking an antiderivative yields the implicit familyOn a regular branch this can also be written $x=\sin y+C/\sin y$. The latter form loses the constant solutions $y=n\pi$, which must be restored separately.

The implicit relation is a first integral of an ordinary differential equation: along any differentiable solution of the original equation,Conversely, a differentiable graph satisfying this relation solves the original equation wherever $\cos y\ne 0$. The original equation is undefined when $\cos y=0$; points at which $2\sin y-x=0$ also need separate attention before solving explicitly for $y^{\prime }$. This is why dividing immediately by $\tan y$ can lose valid solutions and obscure nonuniqueness.