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Given the Differential Equation with the Given Initial Condition y=3t2+12y;y(1)=5y ^ { \prime } = \frac { 3 t ^ { 2 } + 1 } { 2 y } ; \quad y ( 1 ) = - 5

Question 109

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Given the differential equation with the given initial condition: y=3t2+12y;y(1)=5y ^ { \prime } = \frac { 3 t ^ { 2 } + 1 } { 2 y } ; \quad y ( 1 ) = - 5 is this the solution y=t3+t+2?y = - \sqrt { t ^ { 3 } + t + 2 } ?

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