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Chapter 6: Problem 121

If \(y=x+\tan x\), prove that \(\cos ^{2} x \frac{d^{2} y}{d x^{2}}-2 y+2 x=0\)

Short Answer

Expert verified
The derivation and substitution into the differential equation confirms that the given function verifies the differential equation, as the left-hand side equals the right-hand side, \(0 = 0\).

Step by step solution

01

Differentiate the function once

Differentiate the function \(y=x+\tan x\). Using the rules of differentiation, the derivative of \(x\) is \(1\) and the derivative of \(\tan x\) is \(\sec^2 x\). So, \(\frac{dy}{dx}=1+\sec^2 x\)
02

Differentiate the function twice

Next, differentiate the result of step 1 again to get the second derivative. The derivative of \(1\) is \(0\) and the derivative of \(\sec^2 x\) is \(2\sec x * \sec x * \tan x\), using the chain rule. It then simplifies to \(\frac{d^2 y}{dx^2} = 2\sec^2 x \cdot \tan x\)
03

Substitute back in the differential equation

Substitute back \(\frac{dy}{dx}\) and \(\frac{d^2 y}{dx^2}\) in the given differential equation, and simplify. The equation then becomes: \(\cos ^{2} x * 2\sec^2 x \cdot \tan x - 2*(x + \tan x) + 2x\). After some simplification, this becomes \(0 = 0\)

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Most popular questions from this chapter

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

If \(\sqrt{x^{2}+y^{2}}=a e^{\tan ^{-1}} x\), where \(a>0, y(0) \neq 0\) then find the value of \(y^{\prime \prime}(0)\).

\begin{aligned} &\text { If } f(x)=x+\tan x \text { and } g \text { is the inverse of } f \text {, then }\\\ &\text { prove that } g^{\prime}(x)=\frac{1}{2+\tan ^{2}(g(x))} . \end{aligned}

Find \(\frac{d y}{d x}\), if \(y=x^{\sin x}\)

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