Question

If y(t) is a solution of (1+t)dydt−ty=1 and y(0)=−1, then show that y(1)=−12

Answer 1

Answer:

Step-by-step explanation:

If y(t) is a solution of (1+t) dy/dx - ty = 1 and y(0) = -1, then show that y(1) = -1/2 .

Answer 2

$\red\bigstar$QUESTION (CORRECTION):-

If y(t) is a solution of

$\sf (t +1) \dfrac{dy}{dt} - ty = 1 \: and \: y(0) = - 1$

Then show that y(1)=-1/2

$\blue\bigstar$SOLUTION:-

•Given differential equation is

⠀⠀⠀⠀⠀⠀

$\sf \green{\bigstar}(t +1) \dfrac{dy}{dt} - ty = 1 ......(1)$

⠀⠀⠀⠀⠀

$\sf \: or \: \dfrac{dy}{dt} - \bigg( \dfrac{t}{1 + t} \bigg)y = \dfrac{1}{1 + t}$

⠀⠀⠀⠀⠀⠀

$\sf I.F=e^{ \displaystyle \sf \int\dfrac{-t}{1+t}dt}=e^{ \displaystyle \sf \int\bigg(-1+\dfrac{-t}{1+t}dt \bigg)}$

⠀⠀⠀⠀⠀⠀

$\sf = e^{ - t + log(1 + t)}$

⠀⠀⠀⠀⠀⠀

$\sf = e^{ - t} e^{log(1+ t)} = e^{ - t}(1 + t)$

⠀⠀⠀⠀⠀⠀

Hence,the solution of the given differential equation is given by

⠀⠀⠀⠀⠀⠀

$\sf ye^{ - t}(1+ t)= \displaystyle \sf \int \dfrac{{e}^{ - t}(1 + t) }{1 + t} dt$

⠀⠀⠀⠀⠀⠀

$\sf or\:\sf ye^{ - t}(1+ t)= \dfrac{{e}^{ - t} }{ - 1} + C......(2)$

⠀⠀⠀⠀⠀⠀

$\sf As \: \blue{ y(0)=-1},therefore, \pink{y=-1},when \: t=0$

⠀⠀⠀⠀⠀⠀

$\sf - 1 {e}^{0} (1 + 0) = \dfrac{ {e}^{ - 0} }{ - 1} + C$

⠀⠀⠀⠀⠀⠀

$\sf \implies - 1 = - 1 + C$

⠀⠀⠀⠀⠀⠀

$\sf \implies C=0$

⠀⠀⠀⠀⠀⠀

Hence,from (2),we get

⠀⠀⠀⠀⠀⠀

$\sf ye^{ - t}(1+ t)= \dfrac{ {e}^{ - t} }{ - 1}$

⠀⠀⠀⠀⠀⠀

$\sf y = - \dfrac{1}{(1 + t)}$

⠀⠀⠀⠀⠀⠀

$\sf \: When \: t = 1,then \: y=\dfrac{1}{1 + 1} = -\dfrac{1}{2}$

⠀⠀⠀⠀⠀⠀

$\boxed{ \blue { \sf\therefore \: y(1) = - \dfrac{1}{2} }}$