Determine which of the given functions are solutions to the

Determine which of the given functions are solutions to the following differential equation: 2yy\'= y^2+t-1. Circle the function if it is a solution and x-out the function if it is not a solution. Show your work by working out the left and right hand sides and observing whether your results are equal. Y=squareroot -t y= squareroot 2e^t-t y=t y= squareroot 2e^t+t

Solution

Solution is option (2)

Solution of the above equation :

  

Taking square on both sides we get

Y² = 2e^t – t ……………(i)

Differentiating both side with respect to ‘t’ we get

= 2e^t – 1

Substituting the value of e^t from equation (i) we get

= y² +t +1

This is the required solution