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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.

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

Here $M_y-N_x=(6xy+2)-(4xy+1)=2xy+1$, whereas $N=x(2xy+1)$. Thus the ratio in (b) is $1/x$, and one may choose the nonzero integrating factor $\boxed{\mu (x)=x}$ on either half-line. A negative constant multiple on a half-line is harmless.

The multiplied form isHence $x^3{y}^2+x^{2}y=C$, a quadratic equation for $y$. Solving it givesChoose a real branch with positive discriminant to obtain a differentiable graph; a zero discriminant produces a vertical tangent. Although the ratio derivation initially cancels $2xy+1$, the displayed exact differential verifies the multiplying factor directly wherever $x\ne 0$. At $C=0$ the branches are $y=0$ and $y=-1/x$. Only $y=0$ has a finite smooth extension across $x=0$, where the original equation requires $y=0$.

Here $P=y^2$ and $Q={\cos }^{2}x+y\sin 2x$. Their derivatives giveThus on intervals with $\cos x\ne 0$, choose the integrating factor $\mu =e^{\int 2\tan xdx}={\sec }^{2}x$. The multiplied form isConsequentlyThis implicit family includes $y=0$ and the branch $y=-\cot x$ when $C=0$. Where the coefficient of $y^{\prime }$ is nonzero, its differentiated form directly recovers the original equation. A useful equivalent relation is $y\cos x+y^2\sin x=C\cos x$. At $\cos x=0$, the original equation requires $y=0$; continuation there must be checked in that original equation rather than justified by the singular integrating factor. In particular the zero solution extends everywhere, and $y=-\cot x$ extends smoothly across the cosine zeros where it is defined.

Write the equation as $Pdx+Qdy=0$, with $P=3x+y$ and $Q=2x$. Thenso a suitable integrating factor is $\mu =|x{|}^{-1/2}$ on either half-line. For example, on $x>0$ the resulting exact form isHence the solution, written to cover either sign of $x$, isThe original equation remains meaningful at $x=0$, where it requires $y=0$. Only $C=0$ gives a finite differentiable extension across zero, namely $y=-x$.

On $x\ge 1$, an integrating factor is $1/x$. Multiplication gives $(y/x{)}^{\prime }=1$, so the general solution is $y=x^2+Cx$. The initial value $y(1)=1$ forces $C=0$. Therefore

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.

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 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.

Use $u=\cos x$, so $du=-\sin xdx$. Up to an overall minus sign, the differential form becomesSeeking a monomial integrating factor $u^{\alpha }{y}^{\beta }$, equality of mixed derivatives givesThese give $\alpha =1,\beta =0$, so the integrating factor in the original variables isOne may verify exactness directly without relying on invertibility of the cosine substitution:Thus the implicit first integral isThe equivalence with the differential equation holds directly where $\cos x\ne 0$. The zero solution $y=0$ is included at $C=0$. At a zero of $\cos x$, the original graph equation requires $y=0$ for any finite differentiable continuation, so no extra finite graph with nonzero $y$ is created by multiplying by the vanishing factor. If one allows vertical parametrized integral curves of the differential form, $x=\pi /2+n\pi$ is also such a curve and is already contained in the level set $C=0$.

Write the differential form as $Mdx+Ndy$, with the coefficients as printed. A monomial integrating factor can be found from $\mu =x^{\alpha }{y}^{\beta }$. The exactness condition ${\partial }_y(\mu M)={\partial }_x(\mu N)$ givesSolving yields $\alpha =0,\beta =-3$. Thus, on $y\ne 0$,After multiplication, the differential form isHence the implicit solutions areThe branch $\boxed{y=0}$ was excluded by the integrating factor but satisfies the original equation and must be restored. On any graph branch with $y\ne 0$, differentiation of the boxed first integral recovers the original differential equation.

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 $x\ne 0$, 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 $C{e}^{-x}$.

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.

On an interval excluding $t=0$, an integrating factor for this first-order linear differential equation is $t^4$. Multiplying through givesAn antiderivative then gives $t^{4}y=3t^5/5+C$, soThe constant is arbitrary on each connected interval of the domain; the original differential equation is undefined at zero.

Multiplication by the integrating factor and the product rule turn the left side into a single derivative:Using integration by substitution $u=\cos x$ gives $y\sin x=-{\cos }^{3}x/3+C$. Therefore the general solution isThere is one arbitrary constant of integration, as expected for a first-order ordinary differential equation.