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# If y(t) is a solution of $(1+t)\Large\frac{dy}{dt}\normalsize-ty=1$ and $y(0)=-1,$ then show that y(1)=$-\Large \frac{1}{2}$

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• A linear differential equation of the form.
• $\large\frac{dy}{dx}$$+Py=Q has the general solution as ye^{\int pdx}=\int Q.e^{\int pdx}.dx+c Where e ^{\int pdx} is the integrating factor(I.F) • \int e^{-x}dx=-e^{-x}+c Given y(t) is a solution of (1+t)\large\frac{dy}{dt}$$-ty=1$ and $y(0)=-1$
Consider the equation
$(1+t) \large\frac{dy}{dt}$$-ty=1 This can be written as \large\frac{dy}{dt}-\frac{t}{1+t}$$.y=\large\frac{1}{1+t}$
By dividing throughout by $1+t$
Clearly this is a linear differential equation of the form
$\large\frac{dy}{dx}$$+Py=Q Here P=\large\frac{-t}{1+t} and Q=\large\frac{1}{1+t} Let us first the integrating factor I.F e^{\large\int pdx}=e^{\int \Large\frac{-t}{1+t}dt} Consider I=\int \large\frac{-t}{1+t}$$dt$
Add +1 and subtract 1 to the numerator
$-\int \large \frac{t+1-1}{1+t}dt$
On spliting the terms we get,
$-\bigg[\int dt-\int \large\frac{1}{1+t}$$dt\bigg] Step 2: On integrating we get, -[t-\log(1+t)] Now I.F is e ^{\large -t+\log(1+t)}=e^{-t}.e^{\log (1+t)} But e^{\large\log x}$$=x$
Therefore $I.F=e^{-t}(1+t)$
Hence the required solution is
$y e^{\int pdx}=Q . e^{\int pdx}dt +c$
=>$ye^{-t}.(1+t)=\int \large\frac{1}{1+t}.$$e^{-t}(1+t)dt+c$
$ye^{-t}(1+t)=\int e^{-t}.dt$
Step 3:
On integrating we get,
$ye^{-t}(1+t)=-e^{-t}+c$
Dividing throughout by $e^{-t}$ we get
$y(1+t)=-1+ce^{-t}$
It is given $y(0)=-1$
=>When $t=0,y=-1$
Let us substitute the above values of x and y to obtain the value of c
$(-1)(1+0)=-1+ce^0$
But $e^0=1$
We are asked to prove that the value when
$y(1)=-1/2$
Hence $-1=-1+c=>c=0$
Hence the equation is $y(1+t)=-1$
$y(1)=>t=1$
When $t=1$ we know $c=0$
$y(1+1)=-1$
=>$2y=-1$ Therefore $y=\large\frac{-1}{2}$
Hence $y(1)=\large\frac{-1}{2}$ is proved.