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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)=$-\Large \frac{1}{2}$

1 Answer

  • 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}$$-ty=1$
This can be written as
By dividing throughout by $1+t$
Clearly this is a linear differential equation of the form
Here $P=\large\frac{-t}{1+t}$ and $Q=\large\frac{1}{1+t}$
Let us first the integrating factor I.F
$e^{\large\int pdx}=e^{\int \Large\frac{-t}{1+t}dt}$
Consider $ I=\int \large\frac{-t}{1+t} $$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}$$dt\bigg]$
Step 2:
On integrating we get,
Now I.F is
$e ^{\large -t+\log(1+t)}=e^{-t}.e^{\log (1+t)}$
But $ e^{\large\log x}$$=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,
Dividing throughout by $e^{-t}$ we get
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
But $e^0=1$
We are asked to prove that the value when
Hence $-1=-1+c=>c=0$
Hence the equation is $y(1+t)=-1$
When $t=1$ we know $c=0$
=>$2y=-1$ Therefore $y=\large\frac{-1}{2}$
Hence $y(1)=\large\frac{-1}{2}$ is proved.
answered May 8, 2013 by meena.p
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