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Given the Differential Equation with the Given Initial Condition y=t+1y;y(0)=4y ^ { \prime } = \sqrt { \frac { t + 1 } { y } } ; y ( 0 ) = 4

Question 36

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Given the differential equation with the given initial condition: y=t+1y;y(0)=4y ^ { \prime } = \sqrt { \frac { t + 1 } { y } } ; y ( 0 ) = 4 is this the solution y=((t+1)3/2+4)2/3 ? y = \left( ( t + 1 ) ^ { 3 / 2 } + 4 \right) ^ { 2 / 3 } \text { ? }

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