- #1

Another1

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\(\displaystyle y''(x)+y'(x)+F(x)=0\)

Pleas me a idea

Pleas me a idea

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In summary, the given equation can be rewritten as a first degree linear ordinary differential equation by substituting y'(x) with v(x). By finding the general solution to the associated homogeneous equation, and then using a method to find a solution for the entire equation, the general solution for y'(x) is obtained. Finally, y(x) is found by taking the integral of the solution for y'(x).

- #1

Another1

- 40

- 0

\(\displaystyle y''(x)+y'(x)+F(x)=0\)

Pleas me a idea

Pleas me a idea

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- #2

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Let y'(x) = v(x). Then your equation becomes v'(x) + v(x) = -F(x)Another said:\(\displaystyle y''(x)+y'(x)+F(x)=0\)

Pleas me a idea

Now you have a first degree linear ordinary differential equation. You can't write down a final answer, but there is a method by which you can write it out in terms of F(x)...

-Dan

- #3

HallsofIvy

Science Advisor

Homework Helper

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(You could also see that by writing the equation as $v'= \frac{dv}{dt}=-v$ so that $\frac{dv}{v}= -dt$ and integrating: $ln(v)= -t+ C'$ so $v= e^{-t+ C'}= Ce^{-t}$ with $C= e^{C'}$.)

To find a solution to the entire equation, look for a solution of the form $v(t)= u(t)e^{-t}$ where u is a function to be found. $v'= u'e^{-t}- ue^{-t}$ so $v'+ v= u'e^{-t}- ue^{-t}+ ue^{-t}= u'e^{-t}= -F$. Then $u'= =Fe^{t}$ and $u= -\int Fe^{t}dt$.

The general solution to the entire equation is the general solution to the associated homogeneous equation plus that solution to the entire equation:

$v= y'= Ce^{-t}- \left(\int_0^t F(s)e^sds\right)e^{-t}$

Now y is the integral of that.

The order of a linear ODE is determined by the highest derivative present in the equation. For example, if the equation contains a second derivative (y''), it is a second order linear ODE.

The general form of a second order linear ODE is:

a(x)y'' + b(x)y' + c(x)y = g(x)

where a(x), b(x), and c(x) are functions of x and g(x) is the forcing function.

To solve a second order linear ODE with constant coefficients, you can use the method of undetermined coefficients or the method of variation of parameters. Both methods involve finding the complementary solution and particular solution, and then combining them to get the general solution.

Yes, a second order linear ODE can have complex solutions. This occurs when the roots of the characteristic equation are complex numbers. The general solution will involve complex numbers and the real and imaginary parts will represent the real and imaginary solutions, respectively.

To check if your solution to a second order linear ODE is correct, you can plug it back into the original equation and see if it satisfies the equation. You can also check if it satisfies any initial conditions given in the problem. Additionally, you can use a computer program or graphing calculator to graph the solution and see if it matches the given equation and initial conditions.

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