Question

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime \prime}-2 y^{\prime}+5 y=5 x+4 e^{x}(I+\sin 2 x)\)

Short answer

It's a linear non-homogeneous second-order equation. Find complementary \(y = e^{x}(C_1 cos 2x + C_2sin 2x ) +5 x+4 e^{x}(I+ sin 2 x)\)solution

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \[y^{\backslash prime \backslash prime}-2 y^{\backslash prime}+5 y=5 x+4 e^{\backslash sin 2 x} \]The highest-order derivative is a second derivative, and the coefficients of y and its derivatives are constant. Therefore, this is a linear non-homogeneous second-order differential equation.

Find the Complementary Solution ( y_c )

To solve the homogeneous equation \[y^{\backslash prime \backslash prime}-2 y^{\backslash prime}+5 y=0\] , we solve the characteristic equation: \[r^2 - 2r + 5 = 0 \]Solving this quadratic equation, we find the roots: \[r = 1 \backslashpm 2i \] Thus, the complementary solution is: y_c = e^{x}(C_1\backslashcos 2x + C_2\backslashsin 2x)\

Determine the Particular Solution ( y_p )

Since the nonhomogeneous equation is \[y^{\backslash prime \backslash prime}-2 y^{\backslash prime}+5 y=5 x+4 e^{\backslash sin 2 x}(I+\backslashsin 2 x)\], we try a particular solution of the form: Similarly matching nonhomogeneous parts: \(A_1 x + A_2\) can give a suitable guess. Applying a trial solution and solving the forms A and B values:

Form the General Solution

The general solution of the given differential equation combines the complementary solution and particular solution: \[y = y_c + y_p\] \[y = e^{x}(C_1 \backslashcos 2x + C_2\backslashsin 2x) +y_p_particular \]

Key concepts

linear differential equations

In mathematics, differential equations are used to describe various physical phenomena. A linear differential equation is one where the unknown function and its derivatives appear to the power of one (hence, ‘linear’). These equations can be first-order or second-order based on the highest derivative's order. For example, \(y'' - 2y' + 5y = 5x + 4e^x(I + \sin 2x)\) is a second-order linear differential equation because the highest derivative is the second derivative, and all coefficients of \(y\) and its derivatives are constants.
A distinguishing feature of linear differential equations is their applicability in systems that can be expressed linearly. This means the superposition principle holds, where the sum of solutions is also a solution.
Linear differential equations can be categorized into homogeneous and non-homogeneous. The given example is a non-homogeneous linear differential equation since the right-hand side is not zero. The general solution of such an equation comprises two parts: the complementary solution that addresses the homogeneous part, and the particular solution that satisfies the non-homogeneous part.

complementary solution

The complementary solution (\(y_c\)) of a differential equation handles the homogeneous part (i.e., the part where the right-hand side is zero). For the given equation, this is expressed as \(y'' - 2y' + 5y = 0\).
To find the complementary solution, we solve the characteristic equation. Here, we get the characteristic equation by substituting \(y = e^{rt}\), leading to \(r^2 - 2r + 5 = 0\).
Solving this quadratic equation, we get:

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