Question

Use the family in Problem 1 to find a solution of y+y''=0 that satisfies the boundary conditions y(0)=0,y(1)=1.

Second order linear equations
ANSWERED
asked 2021-05-18

Use the family in Problem 1 to find a solution of \(y+y''=0\) that satisfies the boundary conditions \(y(0)=0,y(1)=1.\)

Expert Answers (1)

2021-05-19

We have to find the solution of the given initial value problem, \(y+y''=0, \ y(0)=0,\ y(1)=1\)
The differential equation can be written as, \(\displaystyle{\left({D}^{{2}}+{1}\right)}{y}={0}\ldots.{\left({1}\right)}\)
where \(\displaystyle{D}==\frac{{d}}{{\left.{d}{x}\right.}},{D}^{{2}}==\frac{{{d}^{{2}}}}{{{\left.{d}{x}\right.}^{{2}}}}\)
So the auxiliari equation of (1) is, \(D^2+1=0 \rightarrow D^2=-1 \rightarrow D=+-i\)
The required general solution is, \(\displaystyle{y}={c}{1}{\cos{{x}}}+{c}{2}{\sin{{x}}}\ldots.{\left({2}\right)}\)
Putting \(y=0\) for \(x=0\) in equation (2), we get, \(0=c1\cos0+c2sin0 \rightarrow c1=0\)
And putting \(y=1\) for \(x=1\) in equation (2), we get, \(1=c1\cos1+c2\sin1 \rightarrow c2=\frac{1}{\sin1} [as\ c1=0]\)
Putting the values of c1 and c2 in equation (2), we get, \(\displaystyle{y}={\left(\frac{{1}}{{\sin{{1}}}}\right)}{\sin{{x}}}\)

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