**7. An analytic solution of cubic-quintic Duffing equation**

Let us consider the following ordinary differential equation [36]

$$
\ddot{\boldsymbol{\alpha}} + a\boldsymbol{\alpha} + \beta \mathbf{x}^3 + \gamma \mathbf{x}^5 = \mathbf{0}, \boldsymbol{\alpha} = \mathbf{x}(t), \tag{92}
$$

which is subjected to the following initial conditions

$$\mathbf{x}(\mathbf{0}) = \mathbf{x}\_0 \text{ and } \mathbf{x}'(\mathbf{0}) = \dot{\mathbf{x}}\_0. \tag{93}$$

#### **Theorem 1**.

a. Suppose that *x*<sup>0</sup> 6¼ 0, then the solution of the i.v.p. (92)-(93) is given by

$$\mathbf{x}(t) = \mathbf{x}\_0 \frac{\sqrt{\mathbf{1} + \lambda} v(t)}{\sqrt{\mathbf{1} + \lambda} v^2(t)},\tag{94}$$

where the function *v* � *v t*ð Þ is the solution to the following Duffing equation

$$\begin{cases} \ddot{\boldsymbol{v}} + p\boldsymbol{v} + q\boldsymbol{v}^{3} = \mathbf{0}, \\ \boldsymbol{v}(\mathbf{0}) := \boldsymbol{v}\_{0} = \mathbf{1}, \\ \boldsymbol{v}'(\mathbf{0}) := \dot{\boldsymbol{v}}\_{0} = \frac{\dot{\boldsymbol{x}}\_{0}}{\boldsymbol{\kappa}\_{0}}(\mathbf{1} + \boldsymbol{\lambda}). \end{cases} \tag{95}$$

The values of the coefficients *p* and *q* are given by

$$p = -\frac{\lambda^2 \left(4a + 3\beta \mathbf{x}\_0^2 + 2\gamma \mathbf{x}\_0^4\right) + \lambda \left(3\beta \mathbf{x}\_0^2 + 4\gamma \mathbf{x}\_0^4\right) + 2\gamma \mathbf{x}\_0^4}{2\lambda^2},\tag{96}$$

$$q = -\frac{2\left(\lambda^2 \left(a + \beta \mathbf{x}\_0^2 + \gamma \mathbf{x}\_0^4\right) + \lambda \left(\beta \mathbf{x}\_0^2 + 2\gamma \mathbf{x}\_0^4\right) + \gamma \mathbf{x}\_0^4\right)}{\lambda},\tag{97}$$

and the value of the quantity *λ* is a solution of the cubic

$$-6\dot{\mathbf{x}}\_0^2 \lambda^3 - \left(6\alpha \mathbf{x}\_0^2 + 6\rho \mathbf{x}\_0^4 + 6\rho \mathbf{x}\_0^6\right) \lambda^2 - \left(3\rho \mathbf{x}\_0^4 + 6\rho \mathbf{x}\_0^6\right) \lambda - 2\rho \mathbf{x}\_0^6 = \mathbf{0}.\tag{98}$$

The solution to the the i.v.p. (95) is obtained from the formulas in the first section.

b. Suppose that *x*<sup>0</sup> ¼ 0, in this case, the solution of the i.v.p. (92)-(93) is given by

$$\varkappa(t) = \frac{\sqrt{\mathbf{1} + \lambda} v(t)}{\sqrt{\mathbf{1} + \lambda v^2(t)}},\tag{99}$$

where the function *v* � *v t*ð Þ is the solution of the following Duffing equation

$$\begin{cases} \ddot{\boldsymbol{v}} + p\boldsymbol{v} + q\boldsymbol{v}^{3} = \mathbf{0}, \\ \boldsymbol{v}(\mathbf{0}) := \boldsymbol{v}\_{0} = \mathbf{0}, \\ \boldsymbol{v}'(\mathbf{0}) := \dot{\boldsymbol{v}}\_{0} = \frac{\dot{\boldsymbol{x}}\_{0}}{\sqrt{\mathbf{1} + \boldsymbol{\lambda}}}. \end{cases} \tag{100}$$

The values of the coefficients *p* and *q* are expressed as

$$p = -\frac{2\chi + (3\beta + 4\gamma)\lambda + (4a + 3\beta + 2\gamma)\lambda^2}{2\lambda^2},\tag{101}$$

$$q = -\frac{2\left(\chi + (\beta + 2\gamma)\lambda + (a + \beta + \gamma)\lambda^2\right)}{\lambda},\tag{102}$$

and the value of *λ* is a solution of the cubic

$$(6a+3\emptyset+2\gamma-6\dot{\mathbf{x}}\_0^2)\lambda^3+(6a+6\emptyset+6\gamma)\lambda^2+(3\emptyset+6\gamma)\lambda+2\gamma=0.\tag{103}$$

Note that the solution of the i.v.p. (100) could be obtained from the formulas in the first section.

• Proof: case (a)

Inserting ansatz (94) into Eq. (92) taking the following equation into consideration

$$
\dot{\nu}^2 = \dot{\nu}\_0^2 + p\nu\_0^2 + \frac{q}{2}\nu\_0^4 - p\nu^2 - \frac{q}{2}\nu^4,\tag{104}
$$

and using Eq. (100), we have

$$\sum\_{j=1}^{\mathfrak{g}} H\_j v(t)^j = \mathbf{0},\tag{105}$$

with

$$\begin{aligned} H\_1 &= \frac{\left(-6p\mathbf{x}\_0^2\boldsymbol{\lambda} - 3q\mathbf{x}\_0^2\boldsymbol{\lambda} + 2\mathbf{x}\_0^2\boldsymbol{\alpha} - 6\dot{\mathbf{x}}\_0^2\boldsymbol{\lambda}^3 - 12\dot{\mathbf{x}}\_0^2\boldsymbol{\lambda}^2 - 6\dot{\mathbf{x}}\_0^2\boldsymbol{\lambda} - 2\mathbf{x}\_0^2\right)}{2\mathbf{x}\_0^2}, \\ H\_3 &= -\left(-3p\boldsymbol{\lambda} + q - \mathbf{x}\_0^2\boldsymbol{\rho}\boldsymbol{\lambda} - \mathbf{x}\_0^2\boldsymbol{\rho} - 2a\boldsymbol{\lambda} + \boldsymbol{\lambda}\right), \\ H\_5 &= \frac{1}{2}\left(q\boldsymbol{\lambda} + 2\mathbf{x}\_0^2\boldsymbol{\rho}\boldsymbol{\lambda}^2 + 2\mathbf{x}\_0^2\boldsymbol{\rho}\boldsymbol{\lambda} + 2\mathbf{x}\_0^4\boldsymbol{\rho}\boldsymbol{\lambda}^2 + 4\mathbf{x}\_0^4\boldsymbol{\rho}\boldsymbol{\lambda} + 2\mathbf{x}\_0^4\boldsymbol{\rho} + 2a\boldsymbol{\lambda}^2\right), \end{aligned}$$

where *j* ¼ 1, 3, 5.

Equating the coefficients *H <sup>j</sup>* to zero gives an algebraic system: *H*<sup>1</sup> ¼ 0, *H*<sup>3</sup> ¼ 0, and *H*<sup>5</sup> ¼ 0. Solving *H*<sup>1</sup> ¼ 0 and *H*<sup>3</sup> ¼ 0 will give the values of *p* and *q* that are given in Eqs. (101)-(102). Finally, by inserting the values of *p* and *q* into *H*<sup>1</sup> ¼ 0, we obtain the cubic Eq. (103). Likewise, the case (b) can be proved.

*Analytical Solutions of Some Strong Nonlinear Oscillators DOI: http://dx.doi.org/10.5772/intechopen.97677*
