how do you solve a 2nd order differential equation like this???

Question

d^2y/dx^2+dy/dx+ax/b=0

what if there was no y' term? if it was d^2y/dx^2+ax/b=0

Answer 1

This is a linear differential equation. It can be written as d^2y/dx^2+dy/dx = -ax/b. The solution is composed of 2 parts, the homogeneous solution and the particular solution.

The homegeneous solution is obtained resolving d^2y/dx^2+dy/dx = 0, The solution of this equation depends on the roots of the algebraic equation r^2 + r = 0, whose roots are 0 and -1. So, according to the theory of differential equations, the homegeneous solution, y_h ,is given by
y_h = c1 e^(-x)+ c2 e^(0x) = c1 e^(-x) + c2, where c1 and c2 are constants.

The particular solution has the form y_p = k1x^2 + k2x + k3, because the right hand side is a polynomialof degree 1. We have to determine k1, k2 and k3. Plugging y_p in the equation, we get
2k1 + 2k1x + k2 = -ax/b. So,
k1 = -a/(2b)
2k1 + k2 = 0 => k2 = -2k1 = a/b, and k3 can be anything.

So, our solution is y = y_h + y_p = c1 e^(-x) + c2 -(a/2b)x^2 +(a/b)x + k3. Since c2 and k3 are constants, so is their sum and we simply have y = c e^(-x) -(a/2b)x^2 +(a/b)x + k, where c and k are constant.

This is not the only way to solve this equation. You could put w = dy/dx, and then you get
dw/dx + w = -ax/b. If you multiply both sides by e^x, then, observing that e^(x)(dw/dx + w) = d/dw (w e^x) = get
d/dw (w e^x) = e^x (-ax/b). Now, you integrate, find w, then integrate again ad you get to the same solution of the first method.

Answer 2

Edit: Oh, I just realized that I accidently treated ax/b as ay/b. Thus the following is not correct for your equation, but if it were ay/b, it would be. :)

There are many ways, but the easiest for this is to recognize that it's a homogeneous linear equation and recall that solutions to linear equations are exponentials:

y = e^{rt} for some r

The idea is to plug this "guess" for a solution into the equation, and then figure out what r should be. If you plug in this y you get:

e^{rt}*(r^2 +r +a/b)=0

If you divide by e^{rt} you get the so called characteristic polynomial for the equation:

r^2 + r +a/b = 0

Using the quadratic formula:

r = (-1+/- sqrt(1-4a/b))/2

Now, assuming 1-4a/b is not zero two solutions to the equation are:

e^{(-1+ sqrt(1-4a/b))/2*t}

and

e^{(-1- sqrt(1-4a/b))/2*t}.

If 1-4a/b is zero, then these are actually the same solution, but you can obtain another solution by multiplying the first one by t. Any linear combination of these is also a solution, so the "general solution" is:

y(t) = A*e^{(-1+ sqrt(1-4a/b))/2*t}+B*e^{(-1+/- sqrt(1-4a/b))/2*t}

where A and B are constants. If you have initial conditions, you can use them to solve for A and B. Note that when 1-4a/b is negative, these could be complex valued functions.

Answer 3

it is a non homogeneous linear differential equation of diploma 2. first divide via 2. then rewrite in terms of differential operators: (D^2-a million)y=(x^2)/2 so the answer would be an arbitrary mix of the answer of the corresponding linear homogeneous differential equation: i.e.(D^2-a million)y=0 and the particular answer to the non homogeneous term: (x^2)/2 first discover the roots of the linear homogeneous differential equation, via factoring the differential operators into its factors. for this reason (D-a million)(D+a million) then discover the roots: a million and -a million: so this offers what you may get the particular answer from a itemizing of linearly self sufficient recommendations: Y(complementary)=Ce^x+Be^-x then discover a undeniable answer to the non homogeneous section: because of the fact it is squared it has 3 repeat roots of 0, so it takes the variety: Y(particular)=A+Dx+Ex^2 then you definitely differentiate that 2 situations: Y''=2E then you definitely take that answer and subtract Y(particular) and set it equivalent to the nonhomogenous element: (x^2)/2 2E-A-Dx-Ex^2=(x^2)/2 then you definitely understand that -Ex^2=(x^2)/2, for this reason E=-.5 D=0 because of the fact there is not any x term on the RHS 2(-.5)-A=0...for this reason A=-a million so the best answer is y=Ce^x+B(e^-x) -a million-(.5x^2)

Answer 4

The homogeneous part is
y''+y'=0, which has the solution
y=C1+C2*e^(-x).

Now we need a particular solution for the original equation.
We try something of the form y=Ax^2+Bx+C.
Then y'=2Ax+B and y''=2A, so we want
2A+2Ax+B+ax/b=0.

Thus, 2A+a/b=0, so A=-a/(2b) and
2A+B=0, so B=-2A=a/b.

Hence, the general solution is
y=C1+C2*e^(-x)-ax/(2b)+a/b.

OOPS, that should be -ax^2/(2b)+ax/b