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Given the Differential Equation with the Given Initial Condition dydt=y2lnt;y(1)=13\frac { d y } { d t } = y ^ { 2 } \ln t ; \quad y ( 1 ) = \frac { 1 } { 3 }

question 14

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Given the differential equation with the given initial condition: dydt=y2lnt;y(1)=13\frac { d y } { d t } = y ^ { 2 } \ln t ; \quad y ( 1 ) = \frac { 1 } { 3 } is this the solution y=12tlnt+t?y = \frac { 1 } { 2 - t \ln t + t } ?


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