vela said:You didn't calculate the integrating factor correctly. Recheck the integral.
charmedbeauty said:ohh it should be a half; thanks
so it should be y = y=-6/5x5 - 2C
the answer in my book has 2x5 + Ax2
charmedbeauty said:=e-ln(2) =-2
Bohrok said:This isn't true; take a look at log/exponential rules.
HallsofIvy said:You seem to be making one algebra mistake after another. Yes, the integral of -2/x is -2ln(x) and the exponential of that is [itex]e^{-2\ln(x)}[/itex] but that is NOT [itex]e/(-2\ln(x))[/itex]. I can't imagine where you got that. It is, rather [itex]1/e^{2\ln(x)}[/itex].
But what you should have done first was simplify -2 ln(x). Do you know how to take that -2 'into' the logarithm?
A first order linear ordinary differential equation (ODE) is an equation that relates a function and its derivative in the form of:
a(x)y'(x) + b(x)y(x) = c(x).
Where a(x), b(x), and c(x) are functions of x.
The general solution to a first order linear ODE is in the form of:
y(x) = Ce-∫a(x)dx + ∫c(x)e∫a(x)dxdx.
Where C is an arbitrary constant.
A first order linear ODE has a linear relationship between the function and its derivative, while a non-linear ODE has a non-linear relationship. This means that the coefficients in a linear ODE are constants, while in a non-linear ODE, they may be functions of x.
To solve a first order linear ODE, you can use the method of integrating factors, which involves multiplying both sides of the equation by an integrating factor to make the left side a perfect derivative. You can also use separation of variables, where you separate the variables on each side of the equation and integrate both sides.
First order linear ODE's are commonly used in the fields of physics, engineering, and economics to model various phenomena such as heat transfer, population growth, and electrical circuits. They are also used in various other fields to model real-world situations and make predictions.