Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App