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Solve the given initial-value problem. give the largest interval i over which the solution is defined. x dy dx + y = 6x + 1, y(1) = 8

User Cocotyty
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1 Answer

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x(\mathrm dy)/(\mathrm dx)+y=6x+1

(\mathrm d)/(\mathrm dx)(xy)=6x+1

xy=\displaystyle\int(6x+1)\,\mathrm dx

xy=3x^2+x+C

Given that
y(1)=8, we have


8=3+1+C\implies C=4

so that the particular solution is


xy=3x^2+x+4

y=3x+1+\frac4x

which is valid as long as
x\\eq0, which in turn suggests the largest interval over which the solution may be valid is
(0,\infty).
User Abanoub
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