Verify that tcostt is a solution of the initial value proble

Verify that -tcos(t)-t is a solution of the initial value problem t(dy/dt)= y +(t^2)sin(t) y(pi)=0

ty\'=y+t^2*sin(t) For y=-tcos(t)-t First of all, is y(pi)=0? Yes. Then: y\'= tsin(t)-cos(t)-1 t(tsin(t)-cos(t)-1) ?=? -tcos(t)-t+t^2*sin(t) And one sees that the two sides of the equation are indeed the same, so yes, that is a solution to the IVP.

$Verify that -tcos(t)-t is a solution of the initial value problem t(dy/dt)= y +(t^2)sin(t) y(pi)=0 Solution ty\'=y+t^2*sin(t) For y=-tcos(t)-t First of all, is$

Verify that tcostt is a solution of the initial value proble

Solution

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