Question

verify that y=-tcost-t is a solution of the initial-value problem t(dy/dt)=y+t^2sint, y(pi)=1

Answer 1

ty' = y + t^2 sin t

ty' - y = t^ sin t

$\frac{ty' - y}{t^2} = sin t$

$( \frac{y}{t}) '$ = sin t

y/t = - cos t + C

y = T (C - cos t)

y (π) = π (C + 1) = 0 , C =  -1

y = t (-1 - cos t) =  -t - t cos t

hope this helps

Answer 2

y = -t∙cos(t) 
dy/dt = t∙sin(t) - cos(t) 

t∙[t∙sin(t) - cos(t)] = [-t∙cos(t)] + t²∙sin(t) 
t²∙sin(t) - t∙cos(t) = -t∙cos(t) + t²∙sin(t) 
or
t(dy/dt)=y+t^2*sin(t), y(pi)=0 
tdy/dt - y = t^2sin(t) 
dy/dt - y/t = tsin(t) 

let P(x) = -1/t, Q(x) = tsin(t) 

IF = e^[∫-1/tdt] = e^[-ln(t)] = e^[ln(t)^-1] = t^(-1) 

................tsin(t)dt 
t^(-1)y =∫ --------------- 
...................t 

y/t = ∫ sin(t)dt 
y/t = - cos(t) + C 

when y(π) = 0 

0/π = - cos(π) + C 
0 = - (-1) + C 
0 = 1 + C 
C = -1 

y/t = - cos(t) - 1 
y = -tcos(t) - t 
y = -t(cos(t) + 1) answer//