Question

A special class of first-order linear equations have the form $a(t)y'(t) + a'(t)y(t) = f(t)$, where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $rac{d}{dt}(a(t)y(t)) = a(t)y'(t) + a'(t)y(t) = f(t)$. Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following. $ty'(t) + y = 4 + 2t$, $y(1) = 6$ y(t) = , where t > 0

          A special class of first-order linear equations have the form $a(t)y'(t) + a'(t)y(t) = f(t)$, where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $rac{d}{dt}(a(t)y(t)) = a(t)y'(t) + a'(t)y(t) = f(t)$. Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following.
$ty'(t) + y = 4 + 2t$, $y(1) = 6$
y(t) = , where t > 0

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A special class of first-order linear equations have the form a(t)y'(t) + a'(t)y(t) = f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form racddt(a(t)y(t)) = a(t)y'(t) + a'(t)y(t) = f(t). Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following.
ty'(t) + y = 4 + 2t, y(1) = 6
y(t) = , where t > 0

Added by Larry J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Zhumagali Shomanov

04/25/2023

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In this case, a(t) = 1 and f(t) = 4 + 2t. Show more…

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A special class of first-order linear equations have the form a(t)y'(t) + a(t)y(t) = f(t), where a and f are given functions of t. Notice that the left side of the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following equation: y'(t) + y(t) = 4 + 2t. y(1) = 6. y(t) = , where t > 0.

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