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Solve the initial value problem.\\ $\frac{ds}{dt} = 36t(9t^2 - 7)^3$, $s(1) = 1$\ The solution is $s = $ 

          Solve the initial value problem.\\
$\frac{ds}{dt} = 36t(9t^2 - 7)^3$, $s(1) = 1$\
The solution is $s = $
Solve the initial value problem. (ds)/(dt) = 36t(9t^2 - 7)^3, s(1) = 1The solution is s =

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Solve the initial value problem ds =36t92-7s11 dt Thesolution is s-
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Transcript

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00:01 Hi, from the question given that we need to solve the given initial value problem that is ds by dt is equal to 36t times of 9t square minus 7 the whole cube and s of 1 is equal to 11.
00:16 Now separate the variable by using the variable separable method.
00:19 So ds is equal to 36t 9t square minus 7 the whole cube dt.
00:26 Now integrating on both sides.
00:29 So we have s is equal to integral 36 is a common term t 9t square minus 7 the whole cube.
00:37 Now let u is equal to 9t square minus 7.
00:41 So this implies du is equal to 18t dt.
00:46 So t dt is equal to du by 18 on substituting.
00:52 So we obtain s is equal to 36 times integral this is dt for t dt substitute du by 18 and u cube.
01:05 So for further simplification 36 divided by 18 integral u to the power of 4 divided by 4 u to the power of 4 divided by 4 this is not integral plus the integrating constant c...
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