I have a basic differential equation: [tex]\frac{dy}{dx} = x + y, y(0) = 1[/tex] Now, when I try to solve this by making it exact [tex]\mu \frac{dy}{dx} + \mu y = \mu x[/tex] I get [tex]\mu = e^{-x}[/tex] and solution [tex]-x-1[/tex]. This doesn't satisfy the initial condition. But when I try to solve it as a non-homogenous equation as: [tex]\frac{dy}{dx} + y= x[/tex] I get [tex]y_p = 2e^x, y_c = -x-1[/tex] so [tex]y = 2e^x-x-1[/tex] Which seems to be a correct & full solution. What was I missing in the first try?
You are solving dy/dx + y = x which is wrong since the original equation is dy/dx = x + y, or dy/dx - y = x so now u(x) = e^(-x) and so on and it works out. And yes, the answer is y = 2e^x - x - 1
I assume this was just a typo, but that equation should be [tex]\mu \frac{dy}{dx} - \mu y = \mu x[/tex] Your error: you forgot the constant of integration. You should have obtained [tex]y = -(x+1) + c/\mu = ce^x - (1+x)[/tex] and then solved for the initial condition [itex]y(0)=1[/itex] yielding [itex]c=2[/itex] or [tex]y = 2e^x -(x+1)[/tex] You're nomenclature is backward here. The solution to the homogeneous equation is called the complementary function and is denoted as [itex]y_c[/itex]. The complementary function generally involves arbitrary constants. A solution to the inhomogeneous equation is called a particular function and is denoted as [itex]y_p[/itex]. In this case, the solution to the homogeneous equation [itex]y^\prime-y=0[/itex] is [tex]y_c = ce^x[/tex] where [itex]c[/itex] is an arbitrary constant and [tex]y_p = -(x+1)[/tex] is a particular solution to the inhomogeneous equation. Combining these, [tex]y = ce^x - (1+x)[/tex] which meets the initial conditions when [itex]c=2[/itex].
Oh, that was just a typo. Thanks D_H, it ends up that I've forgotten the integration constant. About naming, take it easy, I'm just a freshman!