Question

2) Find the solution of the following differential equations with given conditions i) $y'' + 5y' + 6y = 0$ ii) $\frac{d^2y}{dx^2} + 6\frac{dy}{dx} + 9y = 0$; $y(0) = 2$, $\frac{dy}{dx}(0) = 0$ iii) $y'' + 2y' + 5y = 0$ iv) $x^2\frac{d^3y}{dx^3} - 2\frac{dy}{dx} = 0$; $y(1) = 2$, $\frac{dy}{dx}(1) = \frac{d^2y}{dx^2}(1) = 0$

          2) Find the solution of the following differential equations with given
conditions
i) $y'' + 5y' + 6y = 0$
ii) $\frac{d^2y}{dx^2} + 6\frac{dy}{dx} + 9y = 0$; $y(0) = 2$, $\frac{dy}{dx}(0) = 0$
iii) $y'' + 2y' + 5y = 0$
iv) $x^2\frac{d^3y}{dx^3} - 2\frac{dy}{dx} = 0$; $y(1) = 2$, $\frac{dy}{dx}(1) = \frac{d^2y}{dx^2}(1) = 0$

2) Find the solution of the following differential equations with given
conditions
i) y” + 5y' + 6y = 0
ii) (d^2y)/(dx^2) + 6(dy)/(dx) + 9y = 0; y(0) = 2, (dy)/(dx)(0) = 0
iii) y” + 2y' + 5y = 0
iv) x^2(d^3y)/(dx^3) - 2(dy)/(dx) = 0; y(1) = 2, (dy)/(dx)(1) = (d^2y)/(dx^2)(1) = 0

Added by Paul J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Darshan Maheshwari

Step 1

To solve this, we can assume a solution of the form y = e^(rt), where r is a constant. Plugging this into the differential equation, we get r^2e^(rt) + 5re^(rt) + 6e^(rt) = 0. Dividing through by e^(rt), we get r^2 + 5r + 6 = 0. This is a quadratic equation, Show more…

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