Ejercicio de ecuaciones lineales (ecuaciones diferenciales)
Quisiera saber cómo se resuelve este ejercicio:
$$\begin{align}&(2xdy/dx+y)*\sqrt(1+x)=1+2x\end{align}$$
Respuesta de Lucas m
1
;)
Hola Mila Cu!
Escribamos la en la forma M(x, y)dx+N(x, y)dy=0 y veamos si es diferencial exacta:
$$\begin{align}&(2x \frac{dy}{dx} +y) \sqrt{1+x}=1+2x\\&\\&2x \frac{dy}{dx}+y=\frac{1+2x}{\sqrt{1+x}}\\&\\&2x \frac{dy}{dx}=\frac{1+2x}{\sqrt{1+x}}-y \\&\\&2xdy=\Bigg [\frac{1+2x}{\sqrt{1+x}}-y \Bigg]dx\\&\\&\Bigg [y-\frac{1+2x}{\sqrt{1+x}} \Bigg]dx+2xdy=0 \ \ \ (ED)\\&\\&M=\Bigg [y-\frac{1+2x}{\sqrt{1+x}} \Bigg] \Rightarrow M_y=1\\&\\&N=2x \Rightarrow N_x=2\\&\\&M_y \neq Nx\\&No \ es \ diferencial \ exacta\\&\\&Factorintegrante :\\&\\&f(x)=e^{\int \frac{M_y-N_x}{N}dx}=e^{\int \frac{1-2}{2x}dx}=e^{\int \frac{-1}{2x}dx}=e^{\frac{-1}{2}lnx}=e^{lnx^{\frac{-1}{2}}}=x^{\frac{-1}{2}}=\frac{1}{\sqrt x}\\&\\&Multiplicando\ \ \ ED·f(x)\\&\\&\frac{1}{\sqrt x}\Bigg [y-\frac{1+2x}{\sqrt{1+x}} \Bigg]dx+2x \frac{1}{\sqrt x}dy=0 \\&\\&\frac{1}{\sqrt x}\Bigg [y-\frac{1+2x}{\sqrt{1+x}} \Bigg]dx+2x^{\frac{1}{2}}dy=0 \\&es \ diferencial \ exacta \Rightarrow \exists \ F(x,y) \ tal \ que\\&\\&F_x=M\\&F_y=N\\&\\&\\&F_y=2x^{\frac{1}{2}} \Rightarrow F(x,y)= \int 2x^{\frac{1}{2}}dy= 2 \sqrt x· y + h(x)\\&\\&F_x=M \ \ derivando \ \ e \ igualando\\&2 \frac{1}{2 \sqrt x}y+h'(x)=\frac{1}{\sqrt x}\Bigg [y-\frac{1+2x}{\sqrt{1+x}} \Bigg] \Rightarrow\\&\\&h'(x)=-\frac{1+2x}{\sqrt x · \sqrt{1+x}}=- \frac{1+2x}{\sqrt{x+x^2}}\\&\\&h(x)= \int- \frac{1+2x}{\sqrt{x+x^2}}dx=\\&\\&x+x^2=t \Rightarrow(1+2x)dx=dt\\&\\&= - \int t^{\frac{-1}{2}}dt=-2t^{\frac{1}{2}}=-2 \sqrt{x+x^2} +C\\&\\&Luego \\&\\&F(x,y)=2 \sqrt x ·y-2 \sqrt{x+x^2} +C\\&\\&\\&\\&\end{align}$$Saludos
;)
;)
