Consider the first order differential equation with initial condition t'+3t2y=2t, y(1)=16 where is a function of...

    

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Consider the first order differential equation with initial condition t'+3t2y=2t, y(1)=16 where is a function of time t. 1 a. Show that y= + C +3 where C is an arbitrary constant, is the general solution to this differential equation. Compute both terms of the LHS in terms of t. ty= 3t2y == = Add the two terms together and show you can simplify the expression to 2t. b. Use the general solution to find the particular solution that satisfies the initial condition y(1) = 16. y = Consider the first order differential equation with initial condition t'+3t2y=2t, y(1)=16 where is a function of time t. 1 a. Show that y= + C +3 where C is an arbitrary constant, is the general solution to this differential equation. Compute both terms of the LHS in terms of t. ty= 3t2y == = Add the two terms together and show you can simplify the expression to 2t. b. Use the general solution to find the particular solution that satisfies the initial condition y(1) = 16. y =

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To solve the given firstorder differential equation and find the particular solution tha... View the full answer