How can I solve this differential equation?

[skysurani]

diff. eq.

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solve the following promblem
Y^(,,,)-3y^(,,)+31y^(,)-37y=0
i let y = e^kx
y'= ke^kx
y''=k^2e^kx
y'''=k^3e^kx

so i got this
k^3e^kx-3(k^2e^kx)+31(ke^kx)-37(e^kx)=0
e^kx(k^3-3K2+31k-37)=0

so,

(k^3-3K2+31k-37)=0

now i have to find (K)


how should i solve for (k) from this equation k^3-3K2+31k-37=0
can i use synthetic division if yes how should i use it or which other method can i use

is this right

i solve the k by synthetic division
5 1 1 -17 -65
5 30 65
1 6 13 0

so the factor is (k-5) (K^2+6k+13)
then i use this equation
(-b+-squrt(b^2-(4ac)))/2a

and got k = -3 +- 2i

and my fianl answer is
y=c1e^5x+c2e^(-3x)cos(2x)+c3e^(-3x)sin(2x)

sorry this is wrong info i type in
please don't do this problem

 

[gtoguy97]

This is not a complete solution. To solve the given equation, you need to use the Quadratic Formula. The quadratic equation is k^3-3K2+31k-37=0. You can then use the Quadratic Formula to find the values of k: k = (-b ± √(b^2 - 4ac))/2a where a = 1, b = -3, and c = -37. You can then substitute the values of k into the original equation to find the solutions.

 

[M98Ranger]

by synthetic division

Solving a differential equation involves finding the function that satisfies the equation. In this case, the differential equation is a third-order linear homogeneous equation. To solve it, you can use the method of undetermined coefficients or the method of variation of parameters.

The method of undetermined coefficients involves assuming a solution of the form y = e^(rx) and finding the values of r that make the equation true. In this case, you would get the characteristic equation r^3 - 3r^2 + 31r - 37 = 0. From this, you can find the values of r and use them to form the general solution y = c1e^(5x) + c2e^(-3x) + c3e^(-3x)cos(2x) + c4e^(-3x)sin(2x).

The method of variation of parameters involves finding a particular solution that satisfies the given equation, and then using it to form the general solution. In this case, you can use the particular solution y = e^(5x) to find the general solution y = c1e^(5x) + c2e^(-3x)cos(2x) + c3e^(-3x)sin(2x).

Both methods involve finding the values of the constants c1, c2, and c3, which can be done by using initial conditions or boundary conditions if they are given.

In summary, solving a differential equation involves finding the general solution that satisfies the given equation, and then using initial or boundary conditions to find the specific solution. It is important to use the appropriate method for the type of differential equation you have.