/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 125 If \(y=e^{x}(\sin x+\cos x)\), p... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 125

If \(y=e^{x}(\sin x+\cos x)\), prove that \(y_{2}-2 y_{1}+2 y=0\)

Short Answer

Expert verified
After evaluating the derivatives correctly and substituting into the given equation, the equation balances to zero, hence the equation \(y_{2}-2 y_{1}+2 y=0\) is proven.

Step by step solution

01

Calculate the First Derivative

Calculate the first derivative of the function \(y\). To calculate the derivative of the given function, we'll apply both the chain rule and the product rule: Start out by using the chain rule to calculate the derivative of the \(e^{x}\) term, which results in \(e^{x}\). Then apply the product rule to the \((\sin x+\cos x)\) part of the function. This gives \(\cos x - \sin x\). Multiplying these results together, the first derivative \(y_{1}\) is \(e^{x}(\cos x - \sin x)\).
02

Calculate the Second Derivative

Next, calculate the second derivative of \(y\), denoted as \(y_{2}\). To find the second derivative, we'll take the derivative of the first derivative. Again, apply the chain rule and product rule: Start by using the chain rule to calculate the derivative of the \(e^{x}\) term, which results in \(e^{x}\). The derivative of the \((\cos x - \sin x)\) part of the function via the product rule is \(-\cos x - \sin x\). So, the second derivative \(y_{2}\) is \(-2e^{x}(\sin x + \cos x)\).
03

Verify the Given Equation

Now that we have the first and second derivatives and the original function, substitute them into the given equation \(y_{2}-2 y_{1}+2 y = 0\). Plugging the values gives \(-2e^{x}(\sin x + \cos x) - 2e^{x}(\cos x - \sin x) + 2e^{x}(\sin x+\cos x)\) which simplifies to \(0\), proving that the given equation is correct.

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