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The differential equation $y^{\prime \prime}+y=0$ has the general solution $y(x)=c_{1} \cos x+c_{2} \sin x$(a) Show that the boundary-value problem $y^{\prime \prime}+y=$$0, \quad y(0)=0, \quad y(\pi)=1$ has no solutions.(b) Show that the boundary-value problem $y^{\prime \prime}+y=$$0, \quad y(0)=0, \quad y(\pi)=0$ has an infinite numberof solutions.

This means that $c_{2}$ is arbitrary, so there is infinitely many solutions to this boundary value problem.

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 2

Basic Ideas and Terminology

Differential Equations

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So in this problem we have a differential equation and a general solution, and we want to check what happens for a couple initial values scenarios. So let's start with the scenario when why of zero is equal to zero and why of pi is equal to one. So let's go ahead and plug it in to our solution here and to our way of X. So let's start with the 1st 1 here. We want y zero equals zero. So zero equal C one times co sign of zero plus C two times Sign of Cyril co sign of zero is one and sign of zero is zero. So this equation gives us that C one is equal to one hurt. Sorry that C one is equal to zero because we have one c one equals zero in this term goes way Now we can check this other initial value. We want one to be equal. Teoh, See one times co sign of pie, Close seed two times sign of pie. So this gives us that one is equal to you. See, one times co sign of pie is negative one plus C two times Sign of pie is zero. So here we get that one is equal to negative C one. But we had that C one is actually equal to zero. So see, one can't be equal to negative one and zero at the same time. So there is no solution to those initial boundary conditions. All right, we can try another set of conditions. So let's try why zero equals zero and why of pie equals zero. In this case, we can plug in this first condition. We want zero equal to see one times co sign of zero plus c Teoh sign of zero. And this of course, gives us again. That zero is equal to see one because co sign of zero is one and sign of zero is zero. On the other hand, why of pi equals zero gives us zero equals. See one times co sign of pie plus C two times sign of pie. Where we have that sign of pi is equal to zero and co sign of pi is equal to negative one like a glove. So this gives us that negative C one equals zero or that c one equals zero. So this means that our particular solution to our differential equation up here is why of axe equals will see one is definitely zero So equal zero plus c two times sign of X where C two can be any real number. So this means that for bees conditions down here, there are infinitely many solutions.

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