How solve this Linear Ordinary Differential Equation?

Question

junyang96 · Answer

$$\frac{ dy }{ dx }+y=e^{-x}$$

junyang96 · Answer

I need a hint

giantrobot · Answer

what do you mean solve? Solve what what? x or y?

giantrobot · Answer

you have to know which value you are solving for before you mess around with the equations.

junyang96 · Answer

Finding the general solutions

mathmale · Answer

$$\frac{ dy }{ dx }+y=e^{-x}$$ displays the derivative of y with respect to x, dy/dx; on the right side, the "forcing function" is a function of x.  This is sufficient info on which to conclude that y is a function of x, and that you are solving for the function y(x).

mathmale · Answer

Since no initial values are given in this problem, you may assume that you are to find a general solution and a particular solution, the latter reflecting the "forcing function" e^(-x).

mathmale · Answer

1) find the general solution of $$\frac{ dy }{ dx }+y=0$$

mathmale · Answer

2) Now let y_p represent the particular solution.  It will have the form y_p=Ae^(-x).  Your job is to find the value of the coefficient, A.

Please share all work you do towards solving this d. e.

junyang96 · Answer

is this the so called complementary solution y=y_0y_p?

mathmale · Answer

What you have just written is the general solution.  It's the sum of the complementary and particular solutions, which can be found separately.

How would you go about finding the complementary solution?

How would you go about finding the particular solution?

junyang96 · Answer

I have used another method to find the general solutions i.e. by multiplying the equation with integrating factor $$I(x)=e^{\int\limits dx}$$

junyang96 · Answer

$$\int\limits \frac{ d }{ dx }(e^{x}y)dx=\int\limits dx$$

junyang96 · Answer

hence, $$y=\frac{x+c}{e^{x}} $$ where c is an arbitrary constant