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Question Number 26904 by sorour87 last updated on 31/Dec/17

$$y^{\left(2\right)}+y=\sec {}^{3}x$$

Commented by prakash jain last updated on 31/Dec/17

$$g\left(x\right)={\sec }^{3}x$$$$\mathrm{Homogenous}\mathrm{solution}$$$${\lambda }^2+1=0\Rightarrow \lambda =\pm i$$$$y_h=c_1{e}^{ix}+c_2{e}^{-ix}$$$$y_h=(c_1+c_2)\cos x+i(c_1-c_2)\cos x$$$$y_h=C_1\cos x+C_2\cos x$$$$y_1\left(x\right)=\cos x$$$$y_2\left(x\right)=\sin \left(x\right)$$$$\mathcal{W}(y_1,y_2)=\left|\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right|=1$$$$u_1\left(x\right)=-\int \frac{y_2\left(x\right)g\left(x\right)}{1}dx$$$$=-\int {\sec }^{3}x\sin xdx=-\int {\sec }^{2}x\tan xdx$$$$=-\frac{1}{2}{\sec }^{2}x$$$$u_2\left(x\right)=\int \frac{y_1\left(x\right)g\left(x\right)}{1}dx$$$$=\int {\sec }^{3}x\cdot \cos xdx=\tan x$$$$y_p\left(x\right)=u_1{y}_1+u_2{y}_2=-\frac{\sec x}{2}+\sin x\tan x$$

Answered by prakash jain last updated on 31/Dec/17

$$y=y_h+y_p$$

Commented by sorour87 last updated on 31/Dec/17

$$ok.thanks$$

Commented by prakash jain last updated on 31/Dec/17

The following page desxribes the method of variation of parameters. http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

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