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# Answer to Question #3275 in Differential Equations for mark kevin

Question #3275
y[sup]2[/sup] e[sup]2x[/sup] + log(3,y) = 2xln y

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Assignment Expert
04.07.11, 12:39

<img src="/cgi-bin/mimetex.cgi?y%5E2%20e%5E%7B2x%7D%20+%20%5Clog_3%20y%20-%202x%20%5Cln%20y%20=%200%20%5C%5C%20F%27_x%20=%202%20y%5E2%20e%5E%7B2x%7D%20-2%20%5Cln%20y%20%5C%5C%20F%27_y%20=%202y%20e%5E%7B2x%7D%20+%20%5Cfrac%7B1%7D%7By%20%5Cln%7B3%7D%7D%20-%20%5Cfrac%7B2x%7D%7By%7D%20%5C%5C%20%5Cfrac%7Bdy%7D%7Bdx%7D%20=%20-%20%5Cfrac%7BF%27_x%7D%7BF%27_y%7D%20=%20-%20%5Cfrac%7B2%20y%5E2%20e%5E%7B2x%7D%20-2%20%5Cln%20y%7D%7B2y%20e%5E%7B2x%7D%20+%20%5Cfrac%7B1%20-2x%20%5Cln%203%7D%7By%20%5Cln%7B3%7D%7D%7D" title="y^2 e^{2x} + \log_3 y - 2x \ln y = 0 \\ F'_x = 2 y^2 e^{2x} -2 \ln y \\ F'_y = 2y e^{2x} + \frac{1}{y \ln{3}} - \frac{2x}{y} \\ \frac{dy}{dx} = - \frac{F'_x}{F'_y} = - \frac{2 y^2 e^{2x} -2 \ln y}{2y e^{2x} + \frac{1 -2x \ln 3}{y \ln{3}}}">

mark kevin
01.07.11, 18:45

y^2 e^2x + log3 y = 2xln y