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Chapter 6: Problem 112

If \(y=c_{1} e^{x}+c_{2} e^{-x}\), then prove that \(\frac{d^{2} y}{d x^{2}}-y=0\).

Short Answer

Expert verified
\(\frac{d^{2}y}{dx^{2}} - y = 0\). The proof is successful.

Step by step solution

01

Differentiate Once

The first step involves taking the derivative of \(y=c_{1} e^{x}+c_{2} e^{-x}\). Using the exponential rule where the derivative of \(e^{ax}\) is \(ae^{ax}\), we find that the derivative is \(\frac{dy}{dx}=c_{1} e^{x}-c_{2} e^{-x}\).
02

Differentiate Twice

The second step is to take the second derivative of our function. It is done by differentiating the derivative function obtained in step 1. By applying the rule again we get \(\frac{d^{2}y}{dx^{2}}=c_{1} e^{x}+c_{2} e^{-x}\).
03

Subtract y from second derivative

As per the problem, we need to then subtract y from the second derivative of y. The equation becomes \(\frac{d^{2}y}{dx^{2}} - y = c_{1} e^{x} + c_{2} e^{-x} - (c_{1} e^{x}+c_{2} e^{-x}) = 0\).

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Most popular questions from this chapter

If \((a+b x) e^{\frac{x}{y}}=x\), then prove that \(x^{3} \frac{d^{2} y}{d x^{2}}=\left(x \frac{d y}{d x}-y\right)^{2}\).

There is a polynomial \(P(x)=a x^{3}+b x^{2}+c x+d\) such that \(P(0)=P(1)=-2, P^{\prime}(0)=-1\), then find the value of \(a+b+c+d+10\)

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If \(y=\sin ^{-1} x\), prove that \(\left(1-x^{2}\right) y_{2}-x y_{1}=0\)

If \(x=a\left(t+\frac{1}{t}\right)\) and \(y=a\left(t-\frac{1}{t}\right)\). then prove that \(\frac{d y}{d x}=\frac{x}{y}\)
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