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Find the particular solution of the differential equation that satisfies the initial condition. Write it out as an equation

Differential Equation: 20xy' - ln(x^5) = 0, x>0
Initial condition: y(1) = 13

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Juan
2 answers

  1. 20xy' - ln(x^5) = 0
    y' = ln(x^5)/(20x) = 1/4 lnx/x
    y = 1/8 (lnx)^2 + C
    since y(1) = 13
    y = 1/8 (lnx)^2 + 13

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    oobleck

  2. Thank you so much that really helps!

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    Juan

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