First order DE with exponential

[dragonblood]

I want to solve a first order differential equation on the form

$$y'+y=e^{x}$$

I want to separate the variables, but not sure how...Thanks!

 

[Mute]

You can't separate the variables directly. The way to solve it is to multiply the whole equation by $e^x$. You'll then notice that since $d (e^x)/dx = e^x$, the right hand side is just the product-rule expansion of $d(e^xy)/dx$. So, you have

$$\frac{d}{dx}(e^x y) = e^{2x}.$$

Now you can separated the variables. You may wonder why I multiplied by e^x. In general you would multiply by some unknown function $\mu(x)$, which you then chose by demanding that you can write the left hand side as $d(\mu y)/dx$. This requires setting the coefficient of y (after multiplication by the $\mu$) equal to $d\mu/dx$. In this case it gives $d\mu/dx = \mu$, which gives the exponential.

In general, the factor that you multiply by will be

$$\mu(x) = \exp\left[\int dx~(\mbox{coefficient of}~y)\right].$$

Notes: this trick only works for first order linear equations, you may not get a $\mu(x)$ which you can express in terms of elementary functions (but it still gives you a solution), and the arbitrary constant of integration in the above expression doesn't matter (it will cancel out in the end).

 

[dragonblood]

Thank you very much!

 

[LCKurtz]

I want to solve a first order differential equation on the form

$$y'+y=e^{x}$$

I want to separate the variables, but not sure how...Thanks!

You have already received one reply using linear methods. An even easier method for this is to observe that it is a constant coefficient equation with characteristic equation r+1=0. So the complementary solution is y_c = Ce^(-x) and you can find a particular solution y_p by undetermined coefficients. Then the general solution is:

y = y_c + y_p