Question
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
Differential Equation: 20xy' - ln(x^5) = 0, x>0
Initial condition: y(1) = 13
Answers
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
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
Thank you so much that really helps!
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