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Resolva o problema de valor inicial (PVI).
\(y'(t) = 3t^7-4t^3\)
\(y(0)=-6\).
xmhintEste é um problema de valor inicial. Primeiro, encontre a solução geral e
depois use a condição inicial dada para determinar a antiderivada particular que
satisfaz a condição inicial.
xmhint\begin{align*} y(t) &= \int (3t^7-4t^3)\; dt \\ y(t) &= \frac {3}{\class{xm-answer}{\cssId{xm-answer-1}{\phantom{\rule{7em}{1.5em}}}}}t^8-\class{xm-answer}{\cssId{xm-answer-2}{\phantom{\rule{7em}{1.5em}}}}+C \\ y(0) = -6 &= \frac {3}{\class{xm-answer}{\cssId{xm-answer-3}{\phantom{\rule{7em}{1.5em}}}}}(0^8)-\class{xm-answer}{\cssId{xm-answer-4}{\phantom{\rule{7em}{1.5em}}}}+C \\ & \Rightarrow C=\class{xm-answer}{\cssId{xm-answer-5}{\phantom{\rule{7em}{1.5em}}}} \\ y(t) &= \frac {3}{\class{xm-answer}{\cssId{xm-answer-6}{\phantom{\rule{7em}{1.5em}}}}}t^8-\left (\class{xm-answer}{\cssId{xm-answer-7}{\phantom{\rule{7em}{1.5em}}}}\right ) \end{align*}
\(y(t) = 21t^6-12t^2-6\) \(y(t) = \frac {3}{8}t^8-t^4-6\) \(y(t) = \frac {3}{8}t^8-t^4\) \(y(t) = \frac {3}{8}t^8-t^4+C\) \(y(t) = 21t^6-12t^2\)