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  • If y(t) is a solution of (1+t)frac{dy}{dt}-ty=1 and y(0) = –1, then show that y(1)=-frac{1}{2}

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

Answers (1)

(1+t)\frac{dy}{dt}-ty=1 and (0,-1) is a solution

To find: Solution for the differential equation

Rewriting the given equation as,

\frac{dy}{dt}-\frac{ty}{1+t}=\frac{1}{1+t}

It is a first order linear differential equation

Comparing it with,

\frac{dy}{dt}-p(t)y=q(t)\\ p(t)=-\frac{t}{1+t}\\ q(t)=\frac{1}{1+t}

Calculation Integrating Factor

IF=e^{\int \frac{t}{1+t}dt}\\ \\ IF=e^{\int \frac{-t+1}{1+t}dt}\\ \\ IF=e^{\int -1+\frac{1}{1+t}dt}\\ \\ IF=e^{-t+\ln(1+t)}=e^{-t}(1+t)\\ \\

Hence the solution for the differential equation is,

y.(IF)=\int q(t).(IF)dt\\ y(e^{-t}(1+t))=\int e^{-t}(1+t)\frac{1}{1+t}dt\\ y(e^{-t}(1+t))=\int e^{-t}dt\\ y(e^{-t}(1+t))=-e^{-t}+c\\ formula: \int e^{-t}dt=-e^{-t}

Substitution (0,-1) to find the value of c
-1=-1+c\\ c=0\\ y(e^{-t}(1+t))=-e^{-t}

The solution therefore y(1) is

y(1)=-\frac{1}{1+1}=-\frac{1}{2}

Posted by

infoexpert24

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