The general first order linear ODE and its integrating factor

$\mathrm{with}\left(\mathrm{DEtools}\right)\:$

$\mathrm{L\_ODE}≔\mathrm{diff}\left(y\left(x\right)\,x\right)=A\left(x\right)y\left(x\right)+B\left(x\right)$

$\mathrm{L\_ODE}≔\frac{\ⅆ}{\ⅆx}y\left(x\right)=A\left(x\right)y\left(x\right)+B\left(x\right)$

$\mathrm{{\rm M}}≔\mathrm{intfactor}\left(\mathrm{L\_ODE}\right)$

$\mathrm{{\rm M}}≔{\ⅇ}^{\int -A\left(x\right)\ⅆx}$

A related first integral can now be obtained either using the option $\mathrm{\_\μ}=\mathrm{{\rm M}}$, as in: firint(L_ODE, _mu = Mu), or multiplying directly L_ODE by Mu (by construction, Mu*L_ODE is exact)

$\mathrm{firint}\left(\mathrm{{\rm M}}\mathrm{L\_ODE}\right)$

${\ⅇ}^{\int -A\left(x\right)\ⅆx}y\left(x\right)-\left(\int {\ⅇ}^{-\left(\int A\left(x\right)\ⅆx\right)}B\left(x\right)\ⅆx\right)+\mathrm{\_C1}=0$

The most general first order ODE reducible by an integrating factor depending only on $x$ is a Linear ODE (see redode):

$\mathrm{reducible\_ODE}≔\mathrm{redode}\left(\mathrm{\mu }\left(x\right)\,y\left(x\right)\right)$

$\mathrm{reducible\_ODE}≔\frac{\ⅆ}{\ⅆx}y\left(x\right)=-\frac{\left(\frac{\ⅆ}{\ⅆx}\mathrm{\mu }\left(x\right)\right)y\left(x\right)+\mathrm{\_F1}\left(x\right)}{\mathrm{\mu }\left(x\right)}$

An exact nonlinear ODE which also has integrating factors of the form $\mathrm{\mu }\left(x\,y\right)$

$\mathrm{ode2}≔xy\left(x\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x{\mathrm{diff}\left(y\left(x\right)\,x\right)}^2+ay\left(x\right)\mathrm{diff}\left(y\left(x\right)\,x\right)+f\left(x\right)$

$\mathrm{ode2}≔xy\left(x\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2+ay\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+f\left(x\right)$

$\mathrm{odeadvisor}\left(\mathrm{ode2}\right)$

$\left[\left[\mathrm{\_2nd\_order}\,\mathrm{\_exact}\,\mathrm{\_nonlinear}\right]\,\left[\mathrm{\_2nd\_order}\,\mathrm{\_reducible}\,\mathrm{\_mu\_xy}\right]\right]$

Two integrating factors and the related first integrals

$\mathrm{\mu }≔\mathrm{intfactor}\left(\mathrm{ode2}\right)$

$\mathrm{\mu }≔1,\frac{x^a}{x}$

$\mathrm{firint}\left(\mathrm{ode2}\right)$

$\frac{{y\left(x\right)}^2\left(a-1\right)}{2}+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)y\left(x\right)+\int f\left(x\right)\ⅆx+\mathrm{\_C1}=0$

$\mathrm{firint}\left(\mathrm{\mu }\left[2\right]\mathrm{ode2}\right)$

$x^{a}y\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+\int \frac{x^{a}f\left(x\right)}{x}\ⅆx+\mathrm{\_C1}=0$

The answer to ode2 can be built from these two first integrals interactively, or by calling dsolve with extra arguments, indicating the use of the integrating factor scheme $\mathrm{mu\_xy}$:

$\mathrm{dsolve}\left(\mathrm{ode2}\,\left['\mathrm{mu\_xy}'\right]\right)$

$y\left(x\right)=\frac{\sqrt{2}\sqrt{\left(a-1\right)\left(x^{1-a}\left(\int \frac{x^{a}f\left(x\right)}{x}\ⅆx\right)+x^{1-a}\mathrm{c\_\_1}-\left(\int f\left(x\right)\ⅆx\right)-\mathrm{c\_\_2}\right)}}{a-1},y\left(x\right)=-\frac{\sqrt{2}\sqrt{\left(a-1\right)\left(x^{1-a}\left(\int \frac{x^{a}f\left(x\right)}{x}\ⅆx\right)+x^{1-a}\mathrm{c\_\_1}-\left(\int f\left(x\right)\ⅆx\right)-\mathrm{c\_\_2}\right)}}{a-1}$

A third order ODE, an integrating factor for it, and the related first integral.

$\mathrm{ode3}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\,x\right)=-\frac{2{\mathrm{diff}\left(y\left(x\right)\,x\right)}^2-y\left(x\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\right)}{{\mathrm{diff}\left(y\left(x\right)\,x\right)}^2\exp \left(\mathrm{diff}\left(y\left(x\right)\,x\,x\right)\right)}$

$\mathrm{ode3}≔\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)=-\frac{2{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2-y\left(x\right)\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)}{{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2{\ⅇ}^{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)}}$

$\mathrm{{\rm M}}≔\mathrm{intfactor}\left(\mathrm{ode3}\right)$

$\mathrm{{\rm M}}≔{\ⅇ}^{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)}$

$\mathrm{firint}\left(\mathrm{{\rm M}}\mathrm{ode3}\right)$

$\frac{y\left(x\right)}{\frac{\ⅆ}{\ⅆx}y\left(x\right)}+{\ⅇ}^{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)}+x+\mathrm{c\_\_1}=0$

The firint command treats linear and nonlinear ODEs in equal footing, unless the optional method = formal, valid only for linear equations, is indicated. Consider

$\mathrm{ode4}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)=\frac{-6x-6}{x\left(3x+2\right)}\mathrm{diff}\left(y\left(x\right)\,x\right)+\frac{6}{x\left(3x+2\right)}y\left(x\right)-\frac{6}{x\left(3x+2\right)}$

$\mathrm{ode4}≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{\left(-6x-6\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{x\left(3x+2\right)}+\frac{6y\left(x\right)}{x\left(3x+2\right)}-\frac{6}{x\left(3x+2\right)}$

Calling firint without optional arguments will result in an error message telling ode4 is not exact. Passing method = formal, you instead get a complete set of independent first integrals obtained by first computing a basis of the solution space

$\mathrm{firint}\left(\mathrm{ode4}\,\mathrm{method}=\mathrm{formal}\right)$

$\frac{\left(-\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)x-\frac{\ⅆ}{\ⅆx}y\left(x\right)+y\left(x\right)-1\right)x^3}{3x+2},\frac{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)x+2y\left(x\right)-2}{3x+2}$

A textbook example: the most general 2nd order non-homogeneous linear ODE having for solution basis $g_1\left(x\right)$ and $g_2\left(x\right)$, for the homogeneous part, and $g_0\left(x\right)$ as particular solution of the whole non-homogeneous equation, so the basis of solutions is $\left[\left[g_1\left(x\right)\,g_2\left(x\right)\right]\,g_0\left(x\right)\right]$

$\mathrm{PDEtools}\left[\mathrm{declare}\right]\left(y\left(x\right)\,g\left(x\right)\,\mathrm{prime}=x\right)$

$y\left(x\right)\mathrm{will\; now\; be\; displayed\; as}y$

$g\left(x\right)\mathrm{will\; now\; be\; displayed\; as}g$

$\mathrm{derivatives\; with\; respect\; to}x\mathrm{of\; functions\; of\; one\; variable\; will\; now\; be\; displayed\; with\; \text{'}}$

$\mathrm{Basis}≔\left[\left[g\left[1\right]\left(x\right)\,g\left[2\right]\left(x\right)\right]\,g\left[0\right]\left(x\right)\right]$

$\mathrm{Basis}≔\left[\left[g_1\,g_2\right]\,g_0\right]$

$\mathrm{ode5}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)=\frac{\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\,x\right)g\left[2\right]\left(x\right)-g\left[1\right]\left(x\right)\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\,x\right)}{g\left[2\right]\left(x\right)\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\right)-g\left[1\right]\left(x\right)\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\right)}\mathrm{diff}\left(y\left(x\right)\,x\right)+\frac{\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\right)\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\,x\right)-\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\,x\right)\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\right)}{g\left[2\right]\left(x\right)\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\right)-g\left[1\right]\left(x\right)\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\right)}y\left(x\right)+\frac{-\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\,x\right)g\left[0\right]\left(x\right)\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\right)+g\left[2\right]\left(x\right)\mathrm{diff}\left(g\left[0\right]\left(x\right)\,x\,x\right)\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\right)-\mathrm{diff}\left(g\left[0\right]\left(x\right)\,x\,x\right)g\left[1\right]\left(x\right)\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\right)+\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\right)g\left[0\right]\left(x\right)\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\,x\right)-g\left[2\right]\left(x\right)\mathrm{diff}\left(g\left[0\right]\left(x\right)\,x\right)\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\,x\right)+\mathrm{diff}\left(g\left[0\right]\left(x\right)\,x\right)g\left[1\right]\left(x\right)\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\,x\right)}{g\left[2\right]\left(x\right)\mathrm{diff}\left(g\left[1\right]\left(x\right)\,x\right)-g\left[1\right]\left(x\right)\mathrm{diff}\left(g\left[2\right]\left(x\right)\,x\right)}$

$\mathrm{ode5}≔\mathrm{y\text{'}\text{'}}=\frac{\left(g_1^{\mathrm{\text{'}\text{'}}}{g}_2-g_1{g}_2^{\mathrm{\text{'}\text{'}}}\right)\mathrm{y\text{'}}}{g_2{g}_1^{\text{'}}-g_1{g}_2^{\text{'}}}+\frac{\left(g_1^{\text{'}}{g}_2^{\mathrm{\text{'}\text{'}}}-g_1^{\mathrm{\text{'}\text{'}}}{g}_2^{\text{'}}\right)y}{g_2{g}_1^{\text{'}}-g_1{g}_2^{\text{'}}}+\frac{-g_2^{\mathrm{\text{'}\text{'}}}{g}_0{g}_1^{\text{'}}+g_2{g}_0^{\mathrm{\text{'}\text{'}}}{g}_1^{\text{'}}-g_0^{\mathrm{\text{'}\text{'}}}{g}_1{g}_2^{\text{'}}+g_2^{\text{'}}{g}_0{g}_1^{\mathrm{\text{'}\text{'}}}-g_2{g}_0^{\text{'}}{g}_1^{\mathrm{\text{'}\text{'}}}+g_0^{\text{'}}{g}_1{g}_2^{\mathrm{\text{'}\text{'}}}}{g_2{g}_1^{\text{'}}-g_1{g}_2^{\text{'}}}$

Verify that Basis is a complete basis for the solutions of ode5: from Basis construct the general solution then test it using odetest

$\mathrm{sol5}≔y\left(x\right)=\mathrm{\_C1}g\left[1\right]\left(x\right)+\mathrm{\_C1}g\left[2\right]\left(x\right)+g\left[0\right]\left(x\right)$

$\mathrm{sol5}≔y=\mathrm{c\_\_1}{g}_1+\mathrm{c\_\_1}{g}_2+g_0$

$\mathrm{odetest}\left(\mathrm{sol5}\,\mathrm{ode5}\right)$

$0$

A complete set of first integrals for ode5 computed from Basis

$\mathrm{firint}\left(\mathrm{ode5}\,y\left(x\right)\,\mathrm{basis}=\mathrm{Basis}\right)$

$\frac{\left(y-g_0\right)g_2^{\text{'}}-g_2\left(\mathrm{y\text{'}}-g_0^{\text{'}}\right)}{g_1{g}_2^{\text{'}}-g_2{g}_1^{\text{'}}},\frac{\left(-y+g_0\right)g_1^{\text{'}}+g_1\left(\mathrm{y\text{'}}-g_0^{\text{'}}\right)}{g_1{g}_2^{\text{'}}-g_2{g}_1^{\text{'}}}$

Verify this result

$\mathrm{map}\left(\mathrm{firtest}\,\left[\right]\,\mathrm{ode5}\,y\left(x\right)\right)$

$\left[0\,0\right]$