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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.

Use the same general solution. The first two conditions now give $A+B=3$ and $-2A-6B+12=-2$, so $A=1$, $B=2$. HenceAt $t=\log \sqrt{2}$, its value is $1/2-3/4+2/8=0$, so the third condition is also satisfied. This is the unique consistent solution.

Put $y=ux$, so $y^{\prime }=u+x{u}^{\prime }$. On an interval with $x\ne 0$, substitution and cancellation giveSince $u^2+2u+2=(u+1{)}^2+1>0$, no constant-$u$ solution is lost when separating. Integrateto obtainAfter restoring $u=y/x$, a local implicit general solution isThe inverse tangent is chosen continuously on a solution branch. Using the continuous polar angle of $(x,x+y)$ extends the expression to charts through $x=0$ when the original equation permits it. A differentiable graph cannot pass through $y=x\ne 0$, because there the original equation would require $5x=0$.

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

Write $f(x,y)=F(u,v)$ in the proposed local coordinates. The multivariable chain rule givesConsequentlyAway from the origin the Jacobian determinant is $u_x{v}_y-u_y{v}_x=4(x^2+y^2)>0$, so the inverse function theorem justifies these local coordinates. Therefore $F_v=0$ and the local general solution isSubstitution verifies the result: $f_x=2x{\Phi }^{\prime }$ and $f_y=-2y{\Phi }^{\prime }$, whose weighted sum vanishes.

There is a global qualification. The coordinate map is two-to-one, and a level set of $x^2-y^2$ has separate characteristic curve branches. A single profile on the entire plane does not follow without a branch-identification assumption. For example, take $f=\exp [-1/(x^2-y^2{)}^2]$ on the wedge $x>0$, $x^2>y^2$, and zero elsewhere. This is smooth, satisfies the equation, and has different values at $(1,0)$ and $(-1,0)$ despite equal $u$. The characteristic components of a hyperbolic first-order equation explain this obstruction. At the origin the coordinate map is singular and the original equation reads $0=0$; the boxed answer is the intended local form on a regular branch, or a global form under an additional identification of branches.

After $n$ integrations, the general solution is an arbitrary polynomial of degree at most $n-1$:Conversely, every such polynomial has zero $n$th derivative. The $n$ free coefficients span the full solution space on a connected interval.

Add the particular solution to the complementary function. The real general solution isThe five constants remain arbitrary until initial data are specified.

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.

Substitution of $y=1$ into the general solution givesThus $F(s)=(s-2)/2$ for $s\ge 2$. Choosing this same formula everywhere gives the polynomial solutionIt directly satisfies the differential equation and $u(x,1)=x^2$.

The boundary condition does not determine a unique solution on the entire plane. Every characteristic ellipse $x^2+2y^2<2$ misses the data line $y=1$. More generally, $F$ is arbitrary for $0\le s<2$, subject to matching the chosen regularity at $s=2$ and at the origin. The displayed polynomial is one global extension; it is forced on $x^2+2y^2\ge 2$. A smooth function supported strictly inside $0<s<2$ may be added to $F$ without changing the boundary data.

Use separation of variables, $f=X(x)T(t)$, and setFor $\kappa =-{\lambda }^2$, with $\lambda >0$, the separated solutions areFor $\kappa =0$, they areFor $\kappa =+{\lambda }^2$, they areArbitrary admissible superpositions of separated solutions are again solutions. An individual product is only a separated family, not the entire general solution. Without prescribed boundaries, the general twice-differentiable solution is equivalently D'Alembert's formula,For example, changing to $\xi =x-ct$, $\eta =x+ct$ makes the differential equation $f_{\xi \eta }=0$, whose two integrations give this form.

The characteristic equation is $(r-1{)}^2=0$, so the repeated-root constant-coefficient differential equation gives the general solution $y=(A+Bx)e^x$. The first boundary condition gives $A=0$, and the second gives $B\pi e^{\pi }=\pi e^{\pi }$, hence $B=1$. Therefore

The nonzero factor $ab/D$ can be cancelled in the local differential equation, leaving $xdy-ydx=0$. On an interval with $x\ne 0$,Hence the explicit general solution ison any interval that avoids the excluded origin. It also follows from fixing the local angle potential, with slope $C=(a/b)\tan (f-C_0)$. As parameterized curves there are also vertical rays $x=0$, which cannot be represented as single-valued functions $y(x)$. Geometrically the connected integral curves are rays, not lines continued through the undefined origin.

The homogeneous characteristic equation is $r^2+2r=r(r+2)=0$, giving the complementary solution $C_1+C_2{e}^{-2x}$. For a particular solution, try $A{x}^2+Bx$. Substitution gives $2A+4Ax+2B=4x$, so $A=1$ and $B=-1$. Therefore the general solution isAs a direct check, the polynomial part contributes $2+2(2x-1)=4x$, while both complementary terms contribute zero.

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 a nonzero separated solution $u=X(x)T(t)$ of the reaction-diffusion equation, substitution and division by $XT$ givesThe two sides depend on different variables, so both equal a separation constant. The homogeneous Dirichlet boundary conditions select a negative spatial constant, written $-k^2$:For $k>0$, $X=A\sin kx+B\cos kx$. The boundary value at zero makes $B=0$; the boundary value at $L$ then requires $kL=n\pi$, with $n=1,2,\dots$. A zero spatial constant gives only a linear function vanishing at both endpoints, hence the zero function; a positive spatial constant similarly gives no nonzero admissible hyperbolic solution. ThereforeBy superposition, the general solution with the stated boundary conditions isThe coefficients are fixed only if initial data are supplied: for $u(x,0)=f(x)$, they are the Fourier sine series coefficients $c_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$. Here they remain arbitrary subject to convergence. Every mode has the ordinary diffusive decay rate plus the additional reaction decay rate one.

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 original PDF has $8y^3-4x$ in the denominator and $dy/dx$ on the left. Rearrange it into the differential formThe cross partial derivatives of its coefficients are both four, so it is an exact differential form. Integrating its first coefficient in $x$ gives a potential $F=\frac{5}{2}{x}^2+4xy+h(y)$. Matching $F_y=4x-8y^3$ gives $h(y)=-2y^4$. Therefore an implicit general solution isDifferentiating verifies $y^{\prime }=(5x+4y)/(8y^3-4x)$. A solution graph is understood only on intervals where its denominator is nonzero; the printed quotient does not define a derivative when $8y^3=4x$.

For the homogeneous linear differential equation, the characteristic equation is $r^2-2r+2=0$, with polynomial roots $1\pm i$. Its homogeneous solution is therefore $e^x(A\cos x+B\sin x)$.

Use the method of undetermined coefficients for a particular solution $y_p=a{x}^2+bx+c$. Substitution givesEquating coefficients yields $a=1$, $b=2$ and $c=1$. Hence the general solution is

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.