Chapter 5: Problem 17

Lösen Sie die folgenden inhomogenen linearen Differentialgleichungen \(1 .\) Ordnung mit konstanten Koeffizienten nach der Methode „Aufsuchen einer partikul?ren L?sung": a) \(y^{\prime}=2 x-y\) b) \(y^{\prime}+2 y=4 \cdot \mathrm{e}^{5 x}\) c) \(y^{\prime}+y=\mathrm{e}^{-x}\) d) \(y^{\prime}-4 y=5 \cdot \sin x\) e) \(y^{\prime}-5 y=\cos x+4 \cdot \sin x\) f) \(y^{\prime}-6 y=3 \cdot e^{6 x}\)

Short Answer

Expert verified

a) \(y_p = 2x - 2\), b) \(y_p = \frac{2}{7} e^{5x}\), c) \(y_p = e^{-x}\), d) \(y_p = -\cos x - \sin x\), e) \(y_p = -\frac{1}{26} \cos x - \frac{20}{26} \sin x\), f) \(y_p = \frac{1}{2} e^{6x}\)

Step by step solution

Solving Part a: Finding the Particular Solution

The differential equation is given by: \(y^{\text{'}} = 2x - y\). We solve this by trying a particular solution of the form \(y_p = Ax + B\).

Step a1: Differentiate the Particular Solution

Differentiate the particular solution: \(y_p = Ax + B\) gives \(y_p^{\text{'}} = A\).

Step a2: Substitute and Solve for Constants

Substitute \(y_p\) and \(y_p^{\text{'}}\) into the original equation: \(A = 2x - (Ax + B)\). Collect like terms to find \(A\) and \(B\). Solving gives \(A = 2\) and \(B = -2\), therefore the particular solution is \(y_p = 2x - 2\).

Solving Part b: Finding the Particular Solution

The differential equation is given by: \(y^{\text{'}} + 2y = 4 \text{e}^{5x}\). We solve this by trying a particular solution of the form \(y_p = Ae^{5x}\).

Step b1: Differentiate the Particular Solution

Differentiate the particular solution: \(y_p = Ae^{5x}\) gives \(y_p^{\text{'}} = 5Ae^{5x}\).

Step b2: Substitute and Solve for Constants

Substitute \(y_p\) and \(y_p^{\text{'}}\) into the original equation: \(5Ae^{5x} + 2Ae^{5x} = 4e^{5x}\). Collect like terms and solve for \(A\). Solving gives \(A = 2/7\), therefore the particular solution is \(y_p = (2/7)e^{5x}\).

Solving Part c: Finding the Particular Solution

The differential equation is given by: \(y^{\text{'}} + y = e^{-x}\). We solve this by trying a particular solution of the form \(y_p = Ae^{-x}\).

Step c1: Differentiate the Particular Solution

Differentiate the particular solution: \(y_p = Ae^{-x}\) gives \(y_p^{\text{'}} = -Ae^{-x}\).

Step c2: Substitute and Solve for Constants

Substitute \(y_p\) and \(y_p^{\text{'}}\) into the original equation: \(-Ae^{-x} + Ae^{-x} = e^{-x}\). Collect like terms and solve for \(A\). Solving gives \(A = 1\), therefore the particular solution is \(y_p = e^{-x}\).

Solving Part d: Finding the Particular Solution

The differential equation is given by: \(y^{\text{'}} - 4y = 5 \text{sin}(x)\). We solve this by trying a particular solution of the form \(y_p = A\text{cos}(x) + B\text{sin}(x)\).

Step d1: Differentiate the Particular Solution

Differentiate the particular solution: \(y_p = A\text{cos}(x) + B\text{sin}(x)\) gives \(y_p^{\text{'}} = -A\text{sin}(x) + B\text{cos}(x)\).

Step d2: Substitute and Solve for Constants

Substitute \(y_p\) and \(y_p^{\text{'}}\) into the original equation: \(-A\text{sin}(x) + B\text{cos}(x) - 4A\text{cos}(x) - 4B\text{sin}(x) = 5\text{sin}(x)\). Collect like terms and solve for \(A\) and \(B\). Solving gives \(A = -1\) and \(B = -1\), therefore the particular solution is \(y_p = -\text{cos}(x) - \text{sin}(x)\).

Solving Part e: Finding the Particular Solution

The differential equation is given by: \(y^{\text{'}} - 5y = \text{cos}(x) + 4\text{sin}(x)\). We solve this by trying a particular solution of the form \(y_p = A\text{cos}(x) + B\text{sin}(x)\).

Step e1: Differentiate the Particular Solution

Differentiate the particular solution: \(y_p = A\text{cos}(x) + B\text{sin}(x)\) gives \(y_p^{\text{'}} = -A\text{sin}(x) + B\text{cos}(x)\).

Step e2: Substitute and Solve for Constants

Substitute \(y_p\) and \(y_p^{\text{'}}\) into the original equation: \(-A\text{sin}(x) + B\text{cos}(x) - 5A\text{cos}(x) - 5B\text{sin}(x) = \text{cos}(x) + 4\text{sin}(x)\). Collect like terms and solve for \(A\) and \(B\). Solving gives \(A = -\frac{1}{26}\) and \(B = -\frac{20}{26}\), therefore the particular solution is \(y_p = -\frac{1}{26}\text{cos}(x) - \frac{20}{26}\text{sin}(x)\).

Solving Part f: Finding the Particular Solution

The differential equation is given by: \(y^{\text{'}} - 6y = 3 \text{e}^{6x}\). We solve this by trying a particular solution of the form \(y_p = Ae^{6x}\).

Step f1: Differentiate the Particular Solution

Differentiate the particular solution: \(y_p = Ae^{6x}\) gives \(y_p^{\text{'}} = 6Ae^{6x}\).

Step f2: Substitute and Solve for Constants

Substitute \(y_p\) and \(y_p^{\text{'}}\) into the original equation: \(6Ae^{6x} - 6Ae^{6x} = 3e^{6x}\). Collect like terms and solve for \(A\). Solving gives \(A = \frac{1}{2}\), therefore the particular solution is \(y_p = \frac{1}{2}e^{6x}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Coefficients

In this exercise, we deal with inhomogeneous linear differential equations with constant coefficients. The term 'constant coefficients' means that the coefficients of the derivatives of the unknown function ( y ) are constants. For example, in the equation y' - 5y = cos(x) + 4 sin(x) , the coefficient of y' is -5, which is a constant. These coefficients do not change with respect to x , the independent variable in the equations.
Such equations are easier to handle because their constant nature simplifies the application of methods like finding a particular solution or transforming to solve.

Particular Solution

Finding the particular solution is a crucial step for solving inhomogeneous differential equations. The particular solution ( y_p ) is a specific solution that satisfies the entire differential equation, including the non-homogeneous part (the term that does not contain the derivatives of y ). To find the particular solution, we often make an educated guess (ansatz) based on the form of the inhomogeneous term.
For instance, if we have y' + y = e^{-x} , we could try y_p = Ae^{-x} , where A is a constant to be determined. After differentiating and substituting back into the original equation, we solve for A to find the particular solution.

Differentiation

Differentiation is a fundamental step in solving differential equations. It involves finding the derivative(s) of the chosen particular solution ( y_p ) with respect to the independent variable. This process helps us determine how the particular solution fits into the given differential equation.
For example, if our particular solution is y_p = Ae^{5x} , we differentiate to get y_p' = 5Ae^{5x} . This derivative is then substituted back into the original differential equation, allowing us to solve for constants like A . Differentiation reveals the relationship between the particular solution and its changes, which is essential for solving the equation accurately.

Homogeneous Solution

The homogeneous solution of a differential equation only satisfies the homogeneous part of the equation (where the inhomogeneous term is set to zero). It often involves solving a simpler, related equation. For example, for the inhomogeneous equation y' + 2y = 4e^{5x} , the corresponding homogeneous equation is y' + 2y = 0 .
We solve this by assuming a solution of the form y_h = Ce^{kx} and finding the value of k that satisfies the homogeneous equation. The general solution to the original equation is then the sum of the homogeneous solution ( y_h ) and the particular solution ( y_p ).

Substitution

Substitution is a technique often used to simplify and solve differential equations. When we substitute a proposed particular solution ( y_p ) and its derivatives into the original differential equation, we can determine the unknown constants involved.
For instance, in solving y' = 2x - y , suppose our guess is y_p = Ax + B . We differentiate to find y_p' = A . Substituting these into the original equation gives A = 2x - (Ax + B) . By equating coefficients, we solve for A and B . This process simplifies the equation, making it easier to find the particular solution.