Question

Determine the general solution of the given differential equation. \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3+\cos 2 t\)

Short answer

The general solution of the given differential equation is \(y(t)=c_1\cos t+c_2\sin t+c_3t\cos t+c_4t\sin t+\frac{1}{13}\cos 2t+3\).

Step-by-step solution

Homogeneous Equation

Let's first consider the homogeneous equation associated with our given differential equation: \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=0\). To solve this, let's consider the characteristic equation: \(r^4+2r^2+1=0\). This factors as \((r^2+1)^2=0\), which gives \(r^2=-1\) with a multiplicity of 2. The roots are \(r=\pm i\) and \(r=\pm i\).

Complementary Function

The roots from the characteristic equation give us the complementary function. Since we have two pairs of complex conjugate roots (\(\pm i\) and \(\pm i\)), the complementary function is of the form: \(y_c(t)=c_1\cos t+c_2\sin t+c_3t\cos t+c_4t\sin t\), where \(c_1\), \(c_2\), \(c_3\), and \(c_4\) are constants.

Trial Function for the Particular Solution

Now we need to find a particular solution for our given non-homogeneous equation, \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3+\cos 2 t\). We can construct a trial function based on the form of our nonhomogeneous term, \(3+\cos 2 t\). Let the trial function be \(y_p(t)=A\cos 2t+B\sin 2t+C\), where \(A\), \(B\), and \(C\) are constants to be determined.

Compute Trial Function Derivatives

We need to compute the first, second, third, and fourth derivatives of our trial function \(y_p(t)=A\cos 2t+B\sin 2t+C\) : \(y_p'(t)=-2A\sin 2t+2B\cos 2t\), \(y_p''(t)=-4A\cos 2t-4B\sin 2t\), \(y_p'''(t)=8A\sin 2t-8B\cos 2t\), \(y_p^{\mathrm{iv}}(t)=16A\cos 2t+16B\sin 2t\).

Substitute Trial Function Derivatives Into the Given Equation

Now we substitute the trial function derivatives into the given differential equation: \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3+\cos 2 t \Rightarrow (16A\cos 2t+16B\sin 2t)+2(-4A\cos 2t-4B\sin 2t)+(A\cos 2t+B\sin 2t+C)=3+\cos 2 t\).

Solve for Constants \(A\), \(B\), and \(C\)

Equating the coefficients of the \(\cos 2t\) and \(\sin 2t\) terms and the constant term on both sides of the equation, we get: \(13A=1\) \(12B=0\) \(C=3\) This gives \(A=\frac{1}{13}\), \(B=0\), \(C=3\).

Particular Solution

Inserting the values for constants \(A\), \(B\), and \(C\) in our trial function, we find the particular solution: \(y_p(t)=\frac{1}{13}\cos 2t+3\).

General Solution

Finally, we can find the general solution by combining the complementary function and the particular solution: \(y(t)=y_c(t)+y_p(t)=(c_1\cos t+c_2\sin t+c_3t\cos t+c_4t\sin t)+(\frac{1}{13}\cos 2t+3)\). So the general solution of the given differential equation is \(y(t)=c_1\cos t+c_2\sin t+c_3t\cos t+c_4t\sin t+\frac{1}{13}\cos 2t+3\).

Key concepts

Homogeneous Equation

A homogeneous equation in differential equations is a type where all terms have the dependent variable or its derivatives, set equal to zero. In this case, we start with the homogeneous equation \(y^{\mathrm{iv}}+2y^{\prime\prime}+y=0\). This step involves stripping away any external functions on the right side of the equation, focusing solely on the intrinsic behavior of the system. The primary aim here is to discover the fundamental form of the solution, which is pivotal in constructing the complete solution later.

This process lays the groundwork for obtaining the complementary function, which represents a part of the solution influenced entirely by the system's internal features.

Characteristic Equation

The characteristic equation is a crucial tool in finding solutions to homogeneous linear differential equations. Derived from the differential equation itself, it translates the problem from a differential form into an algebraic one. By working with the characteristic equation, \(r^4 + 2r^2 + 1 = 0\), we simplify the effort of finding complex solutions into solving for roots of this polynomial.

This equation factors into \((r^2+1)^2=0\), leading us to the roots \(r = \pm i\) with multiplicity. Solving these roots helps us determine the structure of the complementary function, showing the inherent oscillatory nature from the presence of complex roots.

Particular Solution

Finding a particular solution involves addressing the non-homogeneous part of the differential equation. Here, for \(y^{\mathrm{iv}}+2y^{\prime\prime}+y=3+\cos 2t\), the aim is to match the form of the governing non-homogeneous terms. We use a trial function \(y_p(t) = A\cos 2t + B\sin 2t + C\) because it mimics the external influences \(3+\cos 2t\).

After determining the derivatives and substituting back into the equation, equate coefficients to find values for \(A\), \(B\), and \(C\). This gives specific solutions for these constants, culminating in the particular solution \(y_p(t) = \frac{1}{13}\cos 2t + 3\).

This particular solution captures how those outside forces specifically mold the system response.

Complementary Function

The complementary function represents the part of the solution that arises naturally from the system itself. Based on the solutions to the characteristic equation, the complementary function captures oscillatory behavior due to complex roots. For \(r = \pm i\), the form of the complementary function is \(y_c(t) = c_1\cos t + c_2\sin t + c_3t\cos t + c_4t\sin t\).

This function reflects both the inherent oscillations of the system and adjustments due to the repeated roots (indicated by terms multiplying with \(t\)). The constants \(c_1, c_2, c_3,\) and \(c_4\) are arbitrary, representing the unique conditions or initial constraints of the problem.

Blending the complementary function with the particular solution yields the complete expression for tackling initial value problems.