(r-1)^2 * (r^2 2) * (r^2 -2x 2) Find general solution in form: c1e^r1x and so on
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To solve the equation \((r-1)^2 (r^2 + 2) (r^2 - 2x + 2) = 0\) and find the general solution in the form \(c_1 e^{r_1 x} + c_2 e^{r_2 x} + \ldots\), we will follow these steps: Show more…
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Find the general solution of y'' - y = 1 / (e^-x + e^x) y = c1e^x + c2e^-x + (e^x x)/2 + 1/2 sinh x ln(e^2x + 1) y = c1 cos x + c2 sin x + (e^x x)/2 - 1/2 cosh x ln(e^2x + 1) y = c1e^x + c2e^-x + (e^x x)/2 - 1/2 cosh x ln(e^2x + 1) y = c1 cos x + c2 sin x + (e^x x)/2 - 1/2 sinh x ln(e^2x + 1) y = c1e^x + c2e^-x + (e^x x)/2 - 1/2 sinh x ln(e^2x + 1) y = c1e^x + c2e^-x + (e^x x)/2 + 1/2 cosh x ln(e^2x + 1)
Adi S.
Then c2 = -c1 / ((a+2)^2-3). The general pattern, which can be proved by induction, is that cn = -cn-1 / ((a+n)^2-3). [Note: this can be further simplified.] For each of a = -sqrt(3) and a = sqrt(3) we get one solution: y1(x) = c0 * (x^-sqrt(3) - (x^-sqrt(3)+1)/((-sqrt(3)+1)^2-3) + (x^-sqrt(3)+2)/(((-sqrt(3)+2)^2-3)((-sqrt(3)+1)^2-3)) - ...) and y2(x) = c0 * (x^sqrt(3) - (x^sqrt(3)+1)/((sqrt(3)+1)^2-3) + (x^sqrt(3)+2)/(((sqrt(3)+2)^2-3)((sqrt(3)+1)^2-3)) - ...). We can normalize them by setting the first coefficient c0 = 1, because we just want two independent solutions. These two solutions are linearly independent, so every solution is a combination of these two. Any solution is of the form y(x) = a1y1(x) + a2y2(x), for some numbers a1 and a2. Next, compute in the same way the solution for the following equation: x^2y'' + 2xy' - (11x + 1)y = 0 The indicial equation is (use the variable a) = 0. The two values of a are (from lowest to highest) a1 = a2 = A solution corresponding to a1 (with the first coefficient 1) has the power series y1 = x^a1 + x^(a1+1) + x^(a1+2) + .... A solution corresponding to a2 (with the first coefficient 1) has the power series y2 = x^a2 + x^(a2+1) + x^(a2+2) + ....
Sri K.
The general solution of the system x' = (2 2 / 1 3) x is
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