Chapter 2: Q 2.6-42E (page 77)

Question: Use the substitution $y = {vx}^{2}$ to solve
$\frac{dy}{dx} = \frac{2 y}{x} + \cos (\frac{y}{x^{2}})$

Short Answer

Expert verified

$\sec (\frac{y}{x^{2}}) + \tan (\frac{y}{x^{2}}) = C e^{\frac{- 1}{x}}$

Step by step solution

Use the given substitution y=vx2 to solve the given equation

Given equation,

$\frac{dy}{dx} = \frac{2 y}{x} + \cos (\frac{y}{x^{2}})$

Substitute $y = {vx}^{2}$ and $\frac{dy}{dx} = 2 vx + x^{2} \frac{dv}{dx}$ in the above equation,

$2 vx + x^{2} \frac{dv}{dx} = \frac{2 ({vx}^{2})}{x} + \cos (\frac{{vx}^{2}}{x^{2}})$

Simplify the above equation,

$\begin{matrix} 2 vx + x^{2} \frac{dv}{dx} = \frac{2 ({vx}^{2})}{x} + \cos (\frac{{vx}^{2}}{x^{2}}) \\ 2 vx + x^{2} \frac{dv}{dx} = 2 vx + \cos (v) \\ x^{2} \frac{dv}{dx} = \cos (v) \end{matrix}$

Cross multiplication on both sides in the above equation,

$\frac{dv}{\cos (v)} = \frac{dx}{x^{2}}$

Step 2: Integrating

Taking integration of both sides in the above equation

$\int \frac{dv}{\cos (v)} = \int \frac{dx}{x^{2}}$

Solve the above equation,

$\begin{matrix} \int \sec (v) dv = \frac{- 1}{x} + \log C \\ \log (\sec (v) + \tan (v)) = \frac{- 1}{x} + \log C \\ \sec (v) + \tan (v) = C e^{\frac{- 1}{x}} \end{matrix}$

Substitute the value of $v = \frac{y}{x^{2}}$ in the above equation,

$\sec (\frac{y}{x^{2}}) + \tan (\frac{y}{x^{2}}) = C e^{\frac{- 1}{x}}$

Hence the solution is $\sec (\frac{y}{x^{2}}) + \tan (\frac{y}{x^{2}}) = C e^{\frac{- 1}{x}}$

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Question: Use the substitution $y = {vx}^{2}$ to solve
$\frac{dy}{dx} = \frac{2 y}{x} + \cos (\frac{y}{x^{2}})$

Short Answer

Step by step solution

Use the given substitution y=vx2 to solve the given equation

Step 2: Integrating

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Use the given substitution y=vx2 to solve the given equation

Step 2: Integrating

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Question: Use the substitution $y = {vx}^{2}$ to solve
$\frac{dy}{dx} = \frac{2 y}{x} + \cos (\frac{y}{x^{2}})$