From y=c1 e^x cosx+c2 e^x sinx we find y^'=c1 e^x(-sinx+cosx)+c2 e^x(cosx+sinx). (a) We have y(0

step 1 All right, so we have to solve this question. It says, verify that y equals sine x, y equals cos

From y=c1 e^x cosx+c2 e^x sinx we find y^'=c1...

From $y=c_{1} e^{x} \cos x+c_{2} e^{x} \sin x$ we find $y^{\prime}=c_{1} e^{x}(-\sin x+\cos x)+c_{2} e^{x}(\cos x+\sin x).$ (a) We have $y(0)=c_{1}=1, y^{\prime}(0)=c_{1}+c_{2}=0$ so that $c_{1}=1$ and $c_{2}=-1 .$ The solution is $y=e^{x} \cos x-e^{x} \sin x.$ (b) We have $y(0)=c_{1}=1, y(\pi)=-e^{\pi}=-1,$ which is not possible. (c) We have $y(0)=c_{1}=1, y(\pi / 2)=c_{2} e^{\pi / 2}=1$ so that $c_{1}=1$ and $c_{2}=e^{-\pi / 2} .$ The solution is $y=$ $e^{x} \cos x+e^{-\pi / 2} e^{x} \sin x$ (d) We have $y(0)=c_{1}=0, y(\pi)=c_{2} e^{\pi} \sin \pi=0$ so that $c_{1}=0$ and $c_{2}$ is arbitrary. Solutions are $y=c_{2} e^{x} \sin x$ for any real numbers $c_{2}.$

From $y=c_{1} e^{x} \cos x+c_{2} e^{x} \sin x$ we find $y^{\prime}=c_{1} e^{x}(-\sin x+\cos x)+c_{2} e^{x}(\cos x+\sin x).$
(a) We have $y(0)=c_{1}=1, y^{\prime}(0)=c_{1}+c_{2}=0$ so that $c_{1}=1$ and $c_{2}=-1 .$ The solution is $y=e^{x} \cos x-e^{x} \sin x.$
(b) We have $y(0)=c_{1}=1, y(\pi)=-e^{\pi}=-1,$ which is not possible.
(c) We have $y(0)=c_{1}=1, y(\pi / 2)=c_{2} e^{\pi / 2}=1$ so that $c_{1}=1$ and $c_{2}=e^{-\pi / 2} .$ The solution is $y=$ $e^{x} \cos x+e^{-\pi / 2} e^{x} \sin x$
(d) We have $y(0)=c_{1}=0, y(\pi)=c_{2} e^{\pi} \sin \pi=0$ so that $c_{1}=0$ and $c_{2}$ is arbitrary. Solutions are $y=c_{2} e^{x} \sin x$ for any real numbers $c_{2}.$

Key Concepts

Determination of Arbitrary Constants

This concept involves using given conditions, whether initial or boundary values, to solve for the arbitrary constants present in the general solution of a differential equation. By substituting these conditions into the general solution, one can reduce it to a particular solution that satisfies the problem’s constraints.

Boundary Value Problems

Boundary value problems involve conditions imposed at more than one point, typically at the bounds of an interval. These types of problems can be more complex to solve because the conditions must be simultaneously satisfied, and in some cases, they may lead to inconsistencies or non-unique solutions.

Homogeneous Linear Differential Equations

These are differential equations characterized by a linear combination of an unknown function and its derivatives equated to zero. Their solutions form a vector space and are typically expressed as a linear combination of independent functions, each multiplied by an arbitrary constant.

General Solution

The general solution of a differential equation encompasses all possible solutions and includes arbitrary constants. It reflects the structure of the underlying differential equation, and any specific solution can be obtained by assigning particular values to these constants based on additional information.

Initial Value Problems

Initial value problems specify the value of the solution and possibly its derivatives at a single point. These conditions allow for the unique determination of the arbitrary constants in the general solution, leading to a unique solution that fits the initial state of the system.

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