Definición: Se llama ecuación diferencial lineal de primer orden a la que es lineal con respecto a la función (incógnita y) y su derivada.

y'+f(x)y=g(x)

Donde f(x) y g(x) son funciones continuas en un cierto dominio D

En el caso que g(x)=0, se dice que la ecuación lineal es homogénea. Esta ya es de variables separadas.

${\displaystyle \frac{dy}{y}=-f(x)dx}$

Ejemplo: $y\text{'}-2xy=2x{e}^{x^2}$

Resolución: Se hace el cambio y=u(x)v(x)

${\displaystyle u\text{'}v+uv\text{'}+f(x)uv=g(x)\Rightarrow u(v\text{'}-f(x)v)+u\text{'}v=g(x)}$, (*)

v(x) se determina con la condición (v'-f(x)v)=0

Sustituyendo en (*) se obtiene u(x)

Ejemplos:

1. Resolviendo el ejemplo anterior tenemos:

${\displaystyle u\text{'}v+uv\text{'}-2xuv=2x{e}^{x^2}\Rightarrow u(v\text{'}-2xv)+u\text{'}v=2x{e}^{x^2}}$

Haciendo

${\displaystyle \begin{array}{l}v\text{'}-2xv=0\Rightarrow \frac{dv}{v}=2xdx\Rightarrow \ln v=x^2\Rightarrow v=e^{x^2}\\ u\text{'}{e}^{x^2}=2x{e}^{x^2}\Rightarrow \frac{du}{dx}=2x\Rightarrow u=x^2+C\end{array}}$

La integral general

${\displaystyle y=e^{x^2}(x^2+C)}$

2. ${\displaystyle {\left(x-1\right)}^{3}y\text{'}+4{\left(x-1\right)}^{2}y=x+1}$

${\displaystyle \begin{array}{l}u\text{'}v+uv\text{'}+\frac{4uv}{x-1}=\frac{x+1}{{\left(x-1\right)}^3}.\\ u\left(v\text{'}+\frac{4v}{x-1}\right)+u\text{'}v=\frac{x+1}{{\left(x-1\right)}^3}\Rightarrow \\ \Rightarrow \{\begin{array}{l}v\text{'}+\frac{4v}{x-1}=0\Rightarrow \frac{dv}{v}=-4\frac{dx}{x-1}\Rightarrow \ln v=-4\ln (x-1)\Rightarrow v=\frac{1}{{\left(x-1\right)}^4}\\ \frac{1}{{\left(x-1\right)}^4}\frac{du}{dx}=\frac{x+1}{{\left(x-1\right)}^3}\Rightarrow du=(x^2-1)dx\Rightarrow u=\frac{x^3}{3}-x+C\end{array}\end{array}}$

${\displaystyle y=\frac{1}{{\left(x-1\right)}^4}\left(\frac{x^3}{3}-x+C\right)}$

3. ${\displaystyle \mathrm{ch}xdy+(y\mathrm{sh}x+e^x)dx=0}$

${\displaystyle \begin{array}{l}y\text{'}+\frac{\mathrm{sh}x}{\mathrm{ch}x}y+\frac{e^x}{\mathrm{ch}x}=0\Rightarrow u\text{'}v+uv\text{'}+\frac{\mathrm{sh}x}{\mathrm{ch}x}uv+\frac{e^x}{\mathrm{ch}x}=0\\ v\left[u\text{'}+\frac{\mathrm{sh}x}{\mathrm{ch}x}u\right]+uv\text{'}+\frac{e^x}{\mathrm{ch}x}=0\\ u\text{'}+\frac{\mathrm{sh}x}{\mathrm{ch}x}u=0\Rightarrow \frac{du}{u}=-\frac{\mathrm{sh}x}{\mathrm{ch}x}dx\Rightarrow \ln u=-\ln \mathrm{ch}x\Rightarrow u=\frac{1}{\mathrm{ch}x}\\ \frac{1}{\mathrm{ch}x}\frac{dv}{dx}=\frac{-e^x}{\mathrm{ch}x}\Rightarrow v=-e^x+C\end{array}}$

${\displaystyle y=\frac{1}{\mathrm{ch}x}\left(-e^x+C\right)}$

4. ${\displaystyle y\text{'}-\frac{a}{x}y=\frac{x+1}{x},a\in R}$

${\displaystyle \begin{array}{l}u\text{'}v+uv\text{'}-\frac{a}{x}uv=\frac{x+1}{x}\Rightarrow v\left(u\text{'}-\frac{a}{x}u\right)+uv\text{'}=\frac{x+1}{x}.\\ u\text{'}-\frac{a}{x}u=0\Rightarrow \frac{du}{u}=\frac{a}{x}dx\Rightarrow \ln u=a\ln x\Rightarrow u=x^a\\ x^{a}v\text{'}=\frac{x+1}{x}\Rightarrow \frac{dv}{dx}=x^{-(a+1)}(x+1)\Rightarrow v=\frac{x^{-a+1}}{-a+1}+\frac{x^{-a}}{-a}+C\end{array}}$

${\displaystyle y=x^a\left(\frac{x^{-a+1}}{-a+1}+\frac{x^{-a}}{-a}+C\right)=\left(\frac{x}{1-a}-\frac{1}{a}+C{x}^a\right)=\left(\frac{ax+a-1}{a(1-a)}+C{x}^a\right)}$

5. ${\displaystyle y\frac{dx}{dy}+x(y\mathrm{ctg}y+1)=\mathrm{ctg}y}$

${\displaystyle \begin{array}{l}x\text{'}+x\left(\frac{y\mathrm{ctg}y+1}{y}\right)=\frac{\mathrm{ctg}y}{y}\{\begin{array}{l}x=uv\\ u\text{'}v+uv\text{'}+uv\left(\frac{y\mathrm{ctg}y+1}{y}\right)=\frac{\mathrm{ctg}y}{y}\end{array}\\ u\left[v\text{'}+v\left(\frac{y\mathrm{ctg}y+1}{y}\right)\right]+u\text{'}v=\frac{\mathrm{ctg}y}{y}\text{\hspace{0.17em}haciendo\hspace{0.17em}}v\text{'}+v\left(\frac{y\mathrm{ctg}y+1}{y}\right)=0\Rightarrow \\ \Rightarrow \frac{dv}{v}=-\left(\mathrm{ctg}y+\frac{1}{y}\right)dy\Rightarrow \ln v=-\ln \mathrm{sen}y-\ln y\Rightarrow v=\frac{1}{y\mathrm{sen}y}\\ u\text{'}\frac{1}{y\mathrm{sen}y}=\frac{\mathrm{ctg}y}{y}\Rightarrow du=\cos ydy\Rightarrow u=\mathrm{sen}y+C\end{array}}$

${\displaystyle x=\frac{1}{y\mathrm{sen}y}(\mathrm{sen}y+C)=\frac{1}{y}+\frac{C}{y\mathrm{sen}y}}$

6. ${\displaystyle \frac{dy}{dx}+3y=x^2-3x+1}$

${\displaystyle \begin{array}{l}u\text{'}v+uv\text{'}+3uv=x^2-3x+1\Rightarrow v(u\text{'}+3u)+uv\text{'}=x^2-3x+1.\\ u\text{'}+3u=0\Rightarrow \frac{du}{u}=-3dx\Rightarrow \ln u=-3x\Rightarrow u=e^{-3x}\\ e^{-3x}v\text{'}=x^2-3x+1\Rightarrow v={\displaystyle \int e^{3x}(x^2-3x+1)dx=}\\ =\frac{e^{3x}}{3}(x^2-3x+1)-{\displaystyle \int \frac{e^{3x}}{3}(2x-3)dx=}\\ \frac{e^{3x}}{3}(x^2-3x+1)-\frac{e^{3x}}{9}(2x-3)+{\displaystyle \int \frac{e^{3x}}{9}2dx=}\\ =\frac{e^{3x}}{3}(x^2-3x+1)-\frac{e^{3x}}{9}(2x-3)+\frac{2e^{3x}}{27}+C\\ y=e^{-3x}\left[\frac{e^{3x}}{3}(x^2-3x+1)-\frac{e^{3x}}{9}(2x-3)+\frac{2e^{3x}}{27}+C\right]\end{array}}$

${\displaystyle y=\frac{1}{27}\left(9x^2-33x+20\right)+C{e}^{-3x}}$

7. ${\displaystyle (1-x^2)y\text{'}+y-(1-x^2)\sqrt{x+1}=0\Rightarrow y\text{'}+\frac{y}{1-x^2}-\sqrt{x+1}=0}$

${\displaystyle \begin{array}{l}y=u.v\Rightarrow (u\text{'}v+uv\text{'})+\frac{uv}{1-x^2}-\sqrt{x+1}=0\\ v\left(u\text{'}+\frac{u}{1-x^2}\right)+uv\text{'}-\sqrt{x+1}=0\end{array}}$ , (*)

Haciendo

${\displaystyle \begin{array}{l}\left(u\text{'}+\frac{u}{1-x^2}\right)=0\Rightarrow \frac{du}{u}=\frac{dx}{x^2-1}\\ \frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}\Rightarrow \{\begin{array}{l}A=\frac{1}{2}\\ B=\frac{-1}{2}\end{array}\Rightarrow \ln u=\frac{1}{2}\ln \left(\frac{x-1}{x+1}\right)\\ \Rightarrow u=\sqrt{\frac{x-1}{x+1}}\end{array}}$

Volviendo a (*)

${\displaystyle \begin{array}{l}\sqrt{\frac{x-1}{x+1}}\frac{dv}{dx}=\sqrt{x+1}\Rightarrow v={\displaystyle \int \frac{x+1}{\sqrt{x-1}}dx\{\begin{array}{l}x-1=t^2\\ dx=2tdt\end{array}}\\ {\displaystyle \int \frac{t^2+2}{t}2tdt=2{\displaystyle \int (t^2+2}})dt=\frac{2t^3}{3}+4t+C=\frac{2}{3}\sqrt{{\left(x-1\right)}^3}+4\sqrt{x-1}+C\\ y=\sqrt{\frac{x-1}{x+1}}\left(\frac{2}{3}\sqrt{{\left(x-1\right)}^3}+4\sqrt{x-1}+C\right)\end{array}}$

${\displaystyle y=\frac{1}{\sqrt{x+1}}\left(\frac{2}{3}{\left(x-1\right)}^2+4(x-1)+C\sqrt{x-1}\right)}$

Ecuaciones reducibles a lineales

Ecuación de Bernoulli

Es una ecuación de la forma:

${\displaystyle y\text{'}+yf(x)=y^{n}g(x),n\ne 0,1\{\begin{array}{l}n=0\Rightarrow \text{lineal}\\ n=1\Rightarrow \text{variables separadas}\end{array}}$

Resolución: Se divide entre yⁿ, se hace el cambio, $z=\frac{1}{y^{n-1}}$, y ya se tiene la lineal.

Ejemplos:

1. y'-y=xy⁵

${\displaystyle \frac{y\text{'}}{y^5}-\frac{1}{y^4}=x\Rightarrow \{\begin{array}{l}\frac{1}{y^4}=z\\ -4\frac{y\text{'}}{y^5}=z\text{'}\end{array}\Rightarrow -\frac{z\text{'}}{4}-z=x}$

${\displaystyle \begin{array}{l}uv\text{'}+u\text{'}v+4uv=-4x\Rightarrow v(u\text{'}+4u)+uv\text{'}=-4x.\\ u\text{'}+4u=0\Rightarrow \frac{du}{u}=-4dx\Rightarrow \ln u=-4x\Rightarrow u=e^{-4x}\\ e^{-4x}\frac{dv}{dx}=-4x\Rightarrow v=-4{\displaystyle \int x{e}^{4x}dx=-\left(x{e}^{4x}-{\displaystyle \int e^{4x}dx}\right)=e^{4x}\left(\frac{1}{4}-x\right)+C}\end{array}}$

${\displaystyle z=e^{-4x}\left[e^{4x}\left(\frac{1}{4}-x\right)+C\right]=\left(\frac{1}{4}-x\right)+C{e}^{-4x}=\frac{1}{y^4}}$

2. ${\displaystyle y\text{'}+4y=y^2(\mathrm{sen}x+\cos x)}$

Dividiendo entre y² tenemos:

${\displaystyle \frac{y\text{'}}{y^2}+\frac{4}{y}=\mathrm{sen}x+\cos x}$

Haciendo

${\displaystyle \begin{array}{l}\frac{1}{y}=z\Rightarrow z\text{'}=-\frac{y\text{'}}{y^2}\\ -z\text{'}+4z=\mathrm{sen}x+\cos x\end{array}}$

Ecuación lineal. Haciendo z=u·v tenemos

${\displaystyle \begin{array}{l}-(u\text{'}v+uv\text{'})+4uv=\mathrm{sen}x+\cos x.\\ u(-v\text{'}+4v)-u\text{'}v=\mathrm{sen}x+\cos x;haciendo(-v\text{'}+4v)=0\\ \Rightarrow \frac{dv}{v}=4dx\Rightarrow \ln v=4x\Rightarrow v=e^{4x}.\\ -u\text{'}{e}^{4x}=\mathrm{sen}x+\cos x\Rightarrow u=-{\displaystyle \int e^{-4x}(\mathrm{sen}x+\cos x)dx}\end{array}}$

Cíclica

${\displaystyle \begin{array}{l}{\displaystyle \int e^{-4x}(\mathrm{sen}x+\cos x)dx}=-\frac{1}{4}{e}^{-4x}(\mathrm{sen}x+\cos x)+\frac{1}{4}{\displaystyle \int e^{-4x}(\cos x-\mathrm{sen}x)dx=}\\ -\frac{1}{4}{e}^{-4x}(\mathrm{sen}x+\cos x)+\frac{1}{4}\left[-\frac{1}{4}{e}^{-4x}(\cos x-\mathrm{sen}x)+\frac{1}{4}{\displaystyle \int e^{-4x}(-\mathrm{sen}x-\cos x)dx}\right]\Rightarrow \\ \left(1+\frac{1}{16}\right){\displaystyle \int e^{-4x}(\mathrm{sen}x+\cos x)dx}=e^{-4x}\left\{\left(-\frac{1}{4}+\frac{1}{16}\right)\mathrm{sen}x+\left(-\frac{1}{4}-\frac{1}{16}\right)\cos x\right\}\Rightarrow \\ {\displaystyle \int e^{-4x}(\mathrm{sen}x+\cos x)dx}=\frac{16}{17}{e}^{-4x}\left(\frac{-3}{16}\mathrm{sen}x+\frac{-5}{16}\cos x\right)=-\frac{1}{17}{e}^{-4x}(3\mathrm{sen}x+5\cos x)\\ u=\frac{1}{17}{e}^{-4x}(3\mathrm{sen}x+5\cos x)+C\\ \frac{1}{y}=z=uv=\frac{1}{17}\left(3\mathrm{sen}x+5\cos x+17C{e}^{4x}\right)\end{array}}$

${\displaystyle y=\frac{17}{3\mathrm{sen}x+5\cos x+K{e}^{4x}}}$

3. ${\displaystyle y\text{'}=\frac{y(y^2+3x^2)}{2x^3}}$

${\displaystyle 2y\text{'}=\frac{3y}{x}+\frac{y^3}{x^3}\Rightarrow \frac{2y\text{'}}{y^3}=\frac{3}{y^2}\frac{1}{x}+\frac{1}{x^3}}$

con el cambio

${\displaystyle \frac{1}{y^2}=z,\frac{-2y\text{'}}{y^3}=z\text{'}}$

${\displaystyle -z\text{'}=\frac{3z}{x}+\frac{1}{x^3}}$

lineal

${\displaystyle \begin{array}{l}z=uv\Rightarrow -(u\text{'}v+uv\text{'})=\frac{3uv}{x}+\frac{1}{x^3}\\ v(-u\text{'}-\frac{3u}{x})-uv\text{'}=\frac{1}{x^3}\\ -u\text{'}-\frac{3u}{x}=0\Rightarrow \frac{du}{u}=-\frac{3dx}{x}\Rightarrow \ln u=-3\ln x\Rightarrow u=\frac{1}{x^3}\\ -\frac{1}{x^3}\frac{dv}{dx}=\frac{1}{x^3}\Rightarrow dv=-dx\Rightarrow v=-x+C\end{array}}$

${\displaystyle z=\frac{1}{y^2}=\frac{1}{x^3}(-x+C)\Rightarrow x^3=y^2(-x+C)\Rightarrow x(x^2+y^2)=C{y}^2}$

4. ${\displaystyle (x^2+y^2+1)dy+xydx=0}$

${\displaystyle \frac{dx}{dy}+\frac{x}{y}+\frac{1}{x}\left(\frac{y^2+1}{y}\right)=0}$

Multiplicando por x

${\displaystyle \begin{array}{l}xx\text{'}+\frac{x^2}{y}+\left(\frac{y^2+1}{y}\right)=0\\ z=x^2\Rightarrow z\text{'}=2xx\text{'}\end{array}}$

sustituyendo

${\displaystyle \frac{z\text{'}}{2}+\frac{z}{y}+\left(\frac{y^2+1}{y}\right)=0}$

lineal

${\displaystyle \begin{array}{l}\frac{1}{2}(u\text{'}v+uv\text{'})+\frac{uv}{y}+\left(\frac{y^2+1}{y}\right)=0\Rightarrow v\left(\frac{1}{2}u\text{'}+\frac{u}{y}\right)+\frac{1}{2}uv\text{'}+\left(\frac{y^2+1}{y}\right)=0\\ \left(\frac{1}{2}u\text{'}+\frac{u}{y}\right)=0\Rightarrow \frac{du}{u}=-2\frac{dy}{y}\Rightarrow \ln u=-2\ln y\Rightarrow u=\frac{1}{y^2}\\ \frac{1}{2y^2}v\text{'}+\left(\frac{y^2+1}{y}\right)=0\Rightarrow v\text{'}=-2y^2\left(\frac{y^2+1}{y}\right)=-2\left(y^3+y\right)\Rightarrow \\ \Rightarrow v=-2\left(\frac{y^4}{4}+\frac{y^2}{2}\right)+C=-\frac{y^4}{2}-y^2+C\end{array}}$

${\displaystyle z=x^2=\frac{1}{y^2}\left(-\frac{y^4}{2}-y^2+C\right)=-\frac{y^2}{2}-1+\frac{C}{y^2}\Rightarrow 2x^2{y}^2+y^4+2y^2=K}$

5. ${\displaystyle x^{2}y\text{'}+2x^{3}y=y^2(1+2x^2)}$

${\displaystyle \begin{array}{l}y\text{'}+2xy=y^2\left(\frac{1+2x^2}{x^2}\right)\Rightarrow \frac{y\text{'}}{y^2}+2\frac{x}{y}=\left(\frac{1+2x^2}{x^2}\right)\\ z=\frac{1}{y},z\text{'}=-\frac{y\text{'}}{y^2}\Rightarrow -z\text{'}+2zx=\left(\frac{1+2x^2}{x^2}\right)\\ z=uv\Rightarrow -(u\text{'}v+uv\text{'})+2uvx=\left(\frac{1+2x^2}{x^2}\right)\\ u(-v\text{'}+2vx)-u\text{'}v=\left(\frac{1+2x^2}{x^2}\right)=\frac{1}{x^2}+2\\ -v\text{'}+2vx=0\Rightarrow \ln v=x^2\Rightarrow v=e^{x^2}\\ -e^{x^2}u\text{'}=\frac{1}{x^2}+2\Rightarrow u=-{\displaystyle \int e^{-x^2}\left(\frac{1}{x^2}+2\right)dx}\end{array}}$

integrando por partes

${\displaystyle -\left[-\frac{1}{x}{e}^{-x^2}+{\displaystyle \int e^{-x^2}(-2x)}\frac{1}{x}dx+2{\displaystyle \int e^{-x^2}dx}\right]+C=\frac{1}{x}{e}^{-x^2}+C}$

${\displaystyle \frac{1}{y}=e^{x^2}\left(\frac{1}{x}{e}^{-x^2}+C\right)=\frac{1}{x}+e^{x^2}C}$

6. ${\displaystyle \frac{dx}{dy}+\frac{2}{y}x=y^3{x}^3}$

${\displaystyle \begin{array}{l}\frac{x\text{'}}{x^3}+2\frac{1}{x^2}\frac{1}{y}=y^3\Rightarrow \frac{1}{x^2}=z,\frac{-2}{x^3}x\text{'}=z\text{'}\\ -\frac{z\text{'}}{2}+2\frac{z}{y}=y^3\Rightarrow -\frac{1}{2}(u\text{'}v+uv\text{'})+2\frac{uv}{y}=y^3\\ v\left(-\frac{u\text{'}}{2}+\frac{2u}{y}\right)-\frac{uv\text{'}}{2}=y^3;-\frac{u\text{'}}{2}+\frac{2u}{y}=0\Rightarrow \frac{du}{u}=\frac{4}{y}\Rightarrow \\ \ln u=4\ln y\Rightarrow u=y^4.\\ -\frac{y^{4}v\text{'}}{2}=y^3\Rightarrow v=-2{\displaystyle \int \frac{1}{y}dy=-2\ln y+C}\end{array}}$

${\displaystyle \frac{1}{x^2}=y^4(-2\ln y+C)}$

Ecuación de Riccati

Es una ecuación de la forma:

${\displaystyle y\text{'}+f(x)y+g(x)y^2=h(x)}$

Ejemplo

${\displaystyle y\text{'}=1-\frac{1}{2}xy+\frac{1}{2}{y}^2}$

Se observa que y=x es una solución particular de dicha ecuación.

Con el cambio $y=x+\frac{1}{u}$, nos queda

${\displaystyle \begin{array}{l}1-\frac{u\text{'}}{u^2}=1-\frac{1}{2}x\left(x+\frac{1}{u}\right)+\frac{1}{2}{\left(x+\frac{1}{u}\right)}^2\Rightarrow -\frac{u\text{'}}{u^2}=\frac{x}{2u}+\frac{1}{2u^2}\Rightarrow \\ u\text{'}+\frac{1}{2}xu+\frac{1}{2}=0\end{array}}$

lineal

Integrar las ecuaciones:

1. ${\displaystyle (x\ln x)y\text{'}-y=x^3(3\ln x-1)}$

2. ${\displaystyle 2xy\text{'}-y=3x^2}$

3. ${\displaystyle (1+y^2)dx=(\sqrt{1+y^2}\mathrm{sen}y-xy)dy}$

4. ${\displaystyle y\text{'}=\frac{1}{x\mathrm{sen}y+2\mathrm{sen}2y}}$

5. ${\displaystyle x(x^3+1)y\text{'}+(2x^3-1)y=\frac{x^3-2}{x}}$

6. ${\displaystyle 3xy\text{'}-2y=\frac{x^3}{y^2}}$

7. ${\displaystyle y\text{'}=\frac{3x^2}{x^3+y+1}}$

8. ${\displaystyle (2\mathrm{sen}x)\text{}y\text{}\text{'}+y\cos x=y^3(x\cos x-\mathrm{sen}x)}$

9. ${\displaystyle 3xdy=y(1+x\mathrm{sen}x-3y^3\mathrm{sen}x)dx}$

10. ${\displaystyle y\text{'}+\frac{y}{x}+x{y}^2=0}$

11. ${\displaystyle 8xy\text{'}-y=-\frac{1}{y^3\sqrt{x+1}}}$

12. ${\displaystyle y^2(x+x^2)+3+x^{2}yy\text{'}=0}$

13. ${\displaystyle y\text{'}+\frac{y}{x+1}=-\frac{1}{2}{\left(x+1\right)}^3{y}^3}$

14. ${\displaystyle y\text{'}+(1+4x)y-2y^2=2x^2+x+1}$

Solución particular, y=x

15. ${\displaystyle \frac{dy}{dx}+y=y^2(\cos x-\mathrm{sen}x)}$

16. ${\displaystyle xdy-\left[y+x{y}^3(1+\ln x)\right]dx=0}$