Second Order Linear Homogeneous Differential Equations with Constant Coefficients

Solved Problems

Example 3.

Solve the differential equation \[y^{\prime\prime} - 4y' + 5y = 0.\]

Solution.

First we write the corresponding characteristic equation and find its roots:

\[{k^2} - 4k + 5 = 0,\;\; \Rightarrow D = 16 - 4 \cdot 5 = - 4,\;\; \Rightarrow {k_{1,2}} = \frac{{4 \pm \sqrt { - 4} }}{2} = \frac{{4 \pm 2i}}{2} = 2 \pm i.\]

Thus, the differential equation has a pair of complex conjugate roots: \({k_1} = 2 + i,\) \({k_2} = 2 - i.\) In this case, the general solution is expressed by the formula:

\[y\left( x \right) = {e^{2x}}\left[ {{C_1}\cos x + {C_2}\sin x} \right],\]

where \({C_1},\) \({C_2}\) are arbitrary constants.

Example 4.

Solve the equation \[y^{\prime\prime} + 25y = 0.\]

Solution.

The characteristic equation has the form:

\[{k^2} + 25 = 0.\]

This equation has pure imaginary roots:

\[{k^2} = - 25,\;\; \Rightarrow {k_1} = 5i,\;\; {k_2} = - 5i.\]

Then the answer can be written as follows:

\[y\left( x \right) = {C_1}\cos \left( {5x} \right) + {C_2}\sin\left( {5x} \right),\]

where \({C_1},\) \({C_2}\) are constants of integration.

Example 5.

Solve the equation \[y^{\prime\prime} + 4iy = 0.\]

Solution.

In this equation the coefficient before \(y\) is a complex number. The general solution for linear differential equations with constant complex coefficients is constructed in the same way. First we write the characteristic equation:

\[{k^2} + 4i = 0.\]

Determine the roots of the equation:

\[{k^2} = - 4i,\;\; \Rightarrow {k_{1,2}} = \pm \sqrt { - 4i} = \pm \sqrt { - 1} \sqrt 4 \sqrt i = \pm 2i\sqrt i .\]

Calculate separately the square root of the imaginary unit. It is convenient to represent the number \(i\) in trigonometric form:

\[i = \cos \frac{\pi }{2} + i\sin \frac{\pi }{2} = {e^{i \frac{\pi }{2}}},\;\; \Rightarrow \sqrt i = \sqrt {{e^{i \frac{\pi }{2}}}} = e^{\left( {i{\frac{\pi }{2}} \cdot {\frac{1}{2}}} \right)} = e^{i\frac{\pi }{4}} = \cos \frac{\pi }{4} + i\sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2} + i\frac{{\sqrt 2 }}{2}.\]

Then the roots of the characteristic equation are given by

\[{k_{1,2}} = \pm 2i\sqrt i = \pm 2i\left( {\frac{{\sqrt 2 }}{2} + i\frac{{\sqrt 2 }}{2}} \right) = \pm \left( { - \sqrt 2 + \sqrt 2 i} \right),\;\; \Rightarrow {k_1} = - \sqrt 2 + \sqrt 2 i,\;\; {k_2} = \sqrt 2 - \sqrt 2 i.\]

The general solution of the initial differential equation will be expressed through the linear combination of the exponential functions with the found complex numbers:

\[y\left( x \right) = {C_1}{e^{\left( { - \sqrt 2 + \sqrt 2 i} \right)x}} + {C_2}{e^{\left( {\sqrt 2 - \sqrt 2 i} \right)x}},\]

where \({C_1},\) \({C_2}\) are arbitrary constants.