Trigonometry Graph x^2y^24x2y1=0 x2 y2 4x − 2y 1 = 0 x 2 y 2 4 x 2 y 1 = 0 Subtract 1 1 from both sides of the equation x2 y2 4x−2y = −1 x 2 y 2 4 x 2 y = 1Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music2 ( a) Use two power series in x to find the general solution of ( 1 x 2) y ″ 4 x y ′ 2 y = 0 and state the set of x values on which each series solution is valid ( b) Convert the power series
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(x^2+1)y''-4xy'+6y=0 power series
(x^2+1)y''-4xy'+6y=0 power series-(x^2−1) y'' 4xy' 2y = 0 \tag{1} Use the method of grouping to get two product rule derivatives (x^2−1) y'' 4xy' 2y = 0 x^2y''I'm working on finding the general solution of ( x 2 − 1) y ″ 4 x y ′ 2 y = 0 in powers I assume the form y ( x) = ∑ n = 0 ∞ C n x n My basic strategy is to first figure out each piece individually,
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Answer What is the solution to the differential equation (1)?First write the above equation as y'' (4x/x^2 1)y' (2/x^2 1)y =0 Now put y =u and y'' = v Then y'' = v' = (2/x^2 1)y (4x/x^2 1)y' So we get a system of I order equations du/dx = v,About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us Creators
Let the slope of line from (0,0) to the intersection point be m substituting y=mx in xy=2 we get x=\frac {2}{1m} and y=\frac {2m}{1m} substitute in the other equation \frac {4}{(1m)^2} \fracVIDEO ANSWERthe differential equation is why the Monash minus two Y Is equal to zero Using the operator B squared minus 210 That is the characteristic equation is G square minus studentTranscribed image text Solve (x^2 1)y" 4xy' 2y = 2/x 1 such that the general solution obeys the initial conditions y(0) = 1 and y'(0) = 5 and given that y_1 = 1/x 1 and y_2 = 1/x 1 are
Divide \frac{4y}{y1}, the coefficient of the x term, by 2 to get \frac{2y}{y1} Then add the square of \frac{2y}{y1} to both sides of the equation This step makes the left hand side of the
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