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

Answer: (C)

$-\frac {1}{2}$

Answer 2

We begin by determining the integrating factor (IF):

ex: IF =e −∫ t 1+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)=− e 0+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 : $-\frac {1}{2}$