I get a different answer on this separable differential equation than I'm supposed to.

by user773674   Last Updated May 23, 2020 01:20 AM

I let $y = v'$, But when I try and do it i get $v = e^x + e^{c_1} + c_2.$ Can someone please show me? $$v'' -v' =0$$

=> $v=C_1e^x +C_2$, this is supposed to be it.

Answers 2

$$v''-v'=0$$ $$y'-y=0$$ $$(ye^{-x})'=0$$ Integrate $$ye^{-x}=c_1$$ $$v'=c_1e^{x}$$ Integrate again: $$v=c_1e^x+c_2$$

I used integrating factor for solving this DE: $$y'-y=0$$ But it's also separable: $$y'=y$$ $$\dfrac {y'}{y}=1$$ $$\int \dfrac {dy}{y}=\int dx$$ $$\ln y= x+c_1$$ $$y=e^{x+c_1}$$ $$y=ce^x$$

May 22, 2020 17:31 PM

As $$e^{c_1} + c_2 = C(An \ arbitary \ constant)$$ The family of curves is $$ y = e^x + C$$ which satisfies $$y^{\prime} = e^x$$

Dig Amma
Dig Amma
May 22, 2020 17:32 PM

Related Questions

Exact Differential Equation Integrating Factor

Updated February 20, 2019 06:20 AM

Homogeneous 1st Order Differential Equations

Updated February 23, 2019 20:20 PM