7: Handling Higher Order Linear Homogeneous Differential Equations Identical imaginary roots in Java

Generation QR Code in Java 7: Handling Higher Order Linear Homogeneous Differential Equations Identical imaginary roots

Are you curious about the case where the roots of a characteristic equation are duplicate and imaginary For example, take a look at this differential equation with constant coefficients: y(4) + 8y" + 16y = 0 You can try a solution of the following form: y = erx Substituting this solution into your original homogeneous equation gives you: r 4erx + 8r 2erx +16erx = 0 So your characteristic equation is: r 4 + 8r 2 + 16 = 0 You can factor this into: (r 2 + 4) (r 2 + 4) = 0 So the roots are 2i, 2i, 2i, and 2i As you can see, you have duplicate imaginary roots here You can use these equations as I described earlier in this chapter: e( + i )x = e x(cos x + i sin x) and: e( i )x = e x(cos x i sin x) Here are the first two solutions, y1 and y2: y1 = cos 2x and y2 = sin 2x

What about y3 and y4 You can use the same technique as in the previous section: Multiply by progressively higher powers of x until you get all the linearly independent solutions you need In this case, all you need is one power of x to give you: y3 = x cos 2x and y4 = x sin 2x So the general solution to the homogeneous equation is: y = c1 cos 2x + c2 sin 2x + c3x cos 2x + c4x sin 2x

Here s another example; it s a fourth order equation with constant coefficients: y(4) 8y''' + 32y" 64y' + 64y = 0 This equation is no problem You just assume a solution of the form y = erx and plug it into the original equation to get: r 4erx 8r 3erx + 32r 2erx 64rerx + 64erx = 0 So your characteristic equation is: r 4 8r 3 + 32r 2 64r + 64 = 0 Solving this characteristic equation gives you these roots: r1 = 2 + 2i r2 = 2 + 2i r3 = 2 2i r4 = 2 2i So 2 + 2i is a root of multiplicity 2, and so is 2 2i That leads to the following general solution: y = b1e(2 + 2i)x + b2 xe(2 + 2i)x + b3e(2 2i)x + b4 xe(2 2i)x

where b1, b2, b3, and b4 are constants You can rewrite this equation as the following: y = e2x(b1e2ix + b2xe2ix) + e2x(b3e 2ix + b4xe 2ix) or: y = e2x(b1e2ix + b3e 2ix) + e2xx(b2e2ix + b4e 2ix) Once again, you can turn to these relations: ei x = cos x + i sin x and: e i x = cos x i sin x Using these equations gives you: y = e2x(b1(cos 2x + i sin 2x) + b3 (cos 2x - i sin 2x)) + e2xx(b2(cos 2x + i sin 2x) + b4(cos 2x - i sin 2x)) Combining terms and consolidating constants gives you this form for the general solution: y = e2x(c1 cos 2x + c2 sin 2x) + e2xx(c3 cos 2x + c4 sin 2x) And there you go, that s the general solution where the characteristic equation has the roots 2 + 2i, 2 + 2i, 2 2i, and 2 2i

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Breaking down higher order equations with the method of undetermined coefficients Using the variation of parameters to solve higher order equations

he door to your office opens A group of rocket scientists enters, looking embarrassed We need you to solve a differential equation, they say

I thought you were math specialists, you retort We thought so too But this one has us stumped It specifies the shape of the rocket Without it, we can t take off They slide a sheet of paper onto your desk Nobody has to know that we came to you for help, right They look over their shoulders nervously Right, you say, taking a look at the sheet of paper, on which you see this differential equation: y''' + 6y" + 11y' + 6y = 336e5x The equation has the following initial conditions: y(0) = 9 y'(0) = 7 y"(0) = 47