Question

If $y (t)$ is a solution of $(1+t) \frac {dy}{dt}-ty=1 $ and $ y(0)=-1, $ then y(1) is equal to

• A. -0.5
• B. e+1/2 \(e+\frac {1}{2}\)
• C. \(-\frac {1}{2}\)
• D. 44563

Answer 1

The Correct Option is C

We begin by determining the integrating factor (IF):

ex: IF=e−∫t1+t​dt=e−∫t+11+t​dt=e−t+log(1+t)=1+te−t​.

The sought-after solution can be expressed as y(IF)=∫Q⋅IFdt+C, where Q=11+t​ is derived from the given equation.

Thus, y(IF)=∫11+t​⋅1+te−t​dt+C,

=∫e−tdt+C,

=−e−t+C.

Given the initial condition y(0)=−1, we find:

−1⋅(1+0)=−e0+C,

−1=−1+C,

C=0.

Plugging the value of C back in, we have:

y(IF)=−e−t+0,

=−e−t.

Evaluating at t=1:

y(IF)=−e−1,

​\(=-\frac {1}{e}\)

above is example. but here proves that : \(-\frac {1}{2}\)