Question

2. Prove that $y_1 = \cos(\ln x)$ and $y_2 = \sin(\ln x)$ are each solutions of the linear homogeneous equation $x^2y'' + xy' + y = 0$. Then, using the superposition principle, solve the following IVP. $\begin{cases} x^2y'' + xy' + y = 0\\y(1) = -2\\y'(1) = 3 \end{cases}$

          2. Prove that $y_1 = \cos(\ln x)$ and $y_2 = \sin(\ln x)$ are each solutions of the linear homogeneous equation
$x^2y'' + xy' + y = 0$.
Then, using the superposition principle, solve the following IVP.
$\begin{cases} x^2y'' + xy' + y = 0\\y(1) = -2\\y'(1) = 3 \end{cases}$

2. Prove that y1 = cos(ln x) and y2 = sin(ln x) are each solutions of the linear homogeneous equation
x^2y” + xy' + y = 0.
Then, using the superposition principle, solve the following IVP.
x^2y” + xy' + y = 0
y(1) = -2
y'(1) = 3

Added by Michael H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Amy Jiang

08/17/2019

Step 1

To prove that y₁ = cos(lnx) is a solution, we need to find its first and second derivatives and substitute them into the differential equation. y₁ = cos(lnx) y₁' = -sin(lnx) * (1/x) y₁'' = -cos(lnx) * (1/x)² - sin(lnx) * (-1/x²) Substitute y₁, y₁', and y₁'' Show more…

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Prove that y₁ = cos(lnx) and y₂ = sin(lnx) are each solutions of the linear homogeneous equation x²y'' + xy' + y = 0. Then, using the superposition principle, solve the following IVP. x²y'' + xy' + y = 0 y(1) = -2 y'(1) = 3 2. Prove that y₁ = cos(lnx) and y₂ = sin(lnx) are each solutions of the linear homogeneous equation x²y'' + xy' + y = 0. Then, using the superposition principle, solve the following IVP. x²y'' + xy' + y = 0 y(1) = -2 y'(1) = 3