Question #48111
Find one set of values of the numbers a and b, such that the function y = e^(ax) + bln(x) is a solution of the differential equations given below.
a. y^n - ay' = 0
b. y' + y - (b/x) = 0
a. y^n - ay' = 0
b. y' + y - (b/x) = 0
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Comments
Dear Bob. Thank you for correcting us.
for a., why does -a(ae^(ax) + (b/x)) become - a^(2)e^(ax) + b/x? Shouldn't the -a have been multiplied to b/x as well, leaving -a^(2)e^(ax) - a(b/x) instead?