It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 2$ :$q = 5$ :$\map R x = e^{-x} \sec 2 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' + 2 y' + 5 y = 0$ From Linear Second Order ODE: $y'' + 2 y' + 5 y = 0$, this has the general solution: :$y_g = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$ It remains to find a particular solution $y_p$ to $(1)$. Expressing $y_g$ in the form: :$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$ we have: {{begin-eqn}} {{eqn | l = \map {y_1} x       | r = e^{-x} \cos 2 x       | c =  }} {{eqn | l = \map {y_2} x       | r = e^{-x} \sin 2 x       | c =  }} {{eqn | ll= \leadsto       | l = \map { {y_1}'} x       | r = -e^{-x} \cos 2 x - 2 e^{-x} \sin 2 x       | c = Derivative of Exponential Function }} {{eqn | r = -e^{-x} \paren {\cos 2 x + 2 \sin 2 x}       | c =  }} {{eqn | l = \map { {y_2}'} x       | r = -e^{-x} \sin 2 x + 2 e^{-x} \cos 2 x       | c = Derivative of Exponential Function }} {{eqn | r = e^{-x} \paren {2 \cos 2 x - \sin 2 x}       | c =  }} {{end-eqn}} By the Method of Variation of Parameters, we have that: :$y_p = v_1 y_1 + v_2 y_2$ where: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | l = v_2       | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{end-eqn}} where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$. We have that: {{begin-eqn}} {{eqn | l = \map W {y_1, y_2}       | r = y_1 {y_2}' - y_2 {y_1}'       | c = {{Defof|Wronskian}} }} {{eqn | r = e^{-x} \cos 2 x \paren {e^{-x} \paren {2 \cos 2 x - \sin 2 x} } - e^{-x} \sin 2 x \paren {-e^{-x} \paren {\cos 2 x + 2 \sin 2 x} }       | c =  }} {{eqn | r = e^{-2 x} \paren {2 \cos^2 2 x - \cos 2 x \sin 2 x + \sin 2 x \cos 2 x + 2 \sin^2 2 x}       | c =  }} {{eqn | r = 2 e^{-2 x} \paren {\cos^2 2 x + \sin^2 2 x}       | c =  }} {{eqn | r = 2 e^{-2 x}       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v_1       | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int -\frac {e^{-x} \sin 2 x e^{-x} \sec 2 x} {2 e^{-2 x} } \rd x       | c =  }} {{eqn | r = -\frac 1 2 \int \sin 2 x \sec 2 x \rd x       | c =  }} {{eqn | r = -\frac 1 2 \int \frac {\sin 2 x} {\cos 2 x} \rd x       | c = Secant is Reciprocal of Cosine }} {{eqn | r = -\frac 1 2 \int \tan 2 x \rd x       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac 1 4 \ln \cos 2 x       | c = Primitive of $\tan a x$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = v_2       | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x       | c =  }} {{eqn | r = \int \frac {e^{-x} \cos 2 x e^{-x} \sec 2 x} {2 e^{-2 x} } \rd x       | c =  }} {{eqn | r = \frac 1 2 \int \cos 2 x \sec 2 x \rd x       | c =  }} {{eqn | r = \frac 1 2 \int \rd x       | c = Secant is Reciprocal of Cosine }} {{eqn | r = \frac x 2       | c = Primitive of Constant }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = y_p       | r = \paren {\frac 1 4 \ln \cos 2 x} e^{-x} \cos 2 x + \paren {\frac x 2} e^{-x} \sin 2 x       | c =  }} {{eqn | r = \frac {x e^{-x} \sin 2 x} 2 + \frac {e^{-x} \cos 2 x \ln \cos 2 x} 4       | c = rearranging }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \dfrac {x e^{-x} \sin 2 x} 2 + \dfrac {e^{-x} \cos 2 x \ln \cos 2 x} 4$ is the general solution to $(1)$. {{qed}}