Question #149881
Apply the method of variation of parameters to solve the following differential equation
d^2y/dx^2+ x^2/y=secnx , where n is a constant
1
Expert's answer
2020-12-15T19:43:53-0500

d2ydx2+n2y=sec(nx)The solution to the above equation isy=yc+ypwhereycis the complementary factor andyp is the particular integralThe auxiliary equation ism2+n2=0(mjn)(m+jn)=0m=±jn{wherejis a complex number}Recall that if the solution of the auxiliary equationof a second-order differential equation ism=α±jβ,the general solution isy=eαx(C1cosβx+C2sinβx)yc=C1cos(nx)+C2sin(nx)The Wronskian of the two solutions isW(x)=cos(nx)sin(nx)ddx(cos(nx))ddx(sin(nx))=cos(nx)sin(nx)nsin(nx)ncos(nx)=ncos2(nx)+nsin2(nx)=n(cos2(nx)+sin2(nx))=n Our particular solution will be given byyp=V1(x)cos(nx)+V2(x)sin(nx)WhereV1(x)=r(x)sin(nx)W(x)dx,V2(x)=r(x)cos(nx)W(x)dxandV1(x)=r(x)sin(nx)W(x)dx=sec(nx)sin(nx)ndx=tan(nx)ndx=ln(sec(nx))n2+CV2(x)=r(x)cos(nx)W(x)dx=sec(nx)cos(nx)ndx=1ndx=xn+CThe constant terms of the integration can beignored since we are trying to find a non-constantsolution to the differential equationy=yc+yp=C1cos(nx)+C2sin(nx)1n2ln(sec(nx))cos(nx)+xnsin(nx)\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + n^2y = \sec(nx) \\ \textsf{The solution to the above equation is} \\ y = y_c + y_p \\ \textsf{where}\, y_c \, \textsf{is the complementary factor and} \\ y_p\, \textsf{ is the particular integral} \\ \textsf{The auxiliary equation is} \,m^2 + n^2 = 0 \\ \begin{aligned} (m - jn)(m + jn) &= 0 \\ \therefore m &= \pm jn \hspace{1cm} \{\textsf{where}\, j\, \textsf{is a complex number}\} \end{aligned} \\ \textsf{Recall that if the solution of the auxiliary equation}\\ \textsf{of a second-order differential equation is}\\ m = \alpha \pm j\beta, \textsf{the general solution is} \\ y = e^{\alpha x}(C_1\cos{\beta x} + C_2\sin{\beta x})\\ \therefore y_c = C_1\cos(nx) + C_2\sin(nx)\\ \textsf{The Wronskian of the two solutions is} \\ W(x) = \begin{vmatrix} \cos(nx) & \sin(nx)\\ \frac{\mathrm{d}}{\mathrm{d}x}(\cos(nx)) & \frac{\mathrm{d}}{\mathrm{d}x}(\sin(nx)) \end{vmatrix} = \begin{vmatrix} \cos(nx) & \sin(nx)\\ -n\sin(nx) & n\cos(nx) \end{vmatrix} \\ \begin{aligned} &= n\cos^2(nx) + n\sin^2(nx)\\ &= n(\cos^2(nx) + \sin^2(nx)) \\ &= n \end{aligned}\\ \therefore\textsf{ Our particular solution will be given by}\\ y_p = V_1(x)\cos(nx) + V_2(x)\sin(nx)\\ \textsf{Where}\, V_1(x) = -\int \frac{r(x)\sin(nx)}{W(x)} \, \mathrm{d}x,\, V_2(x) = \int \frac{r(x)\cos(nx)}{W(x)}\, \mathrm{d}x \, \textsf{and} \, \\ \begin{aligned} V_1(x) &= -\int \frac{r(x)\sin(nx)}{W(x)} \, \mathrm{d}x \\ &= -\int \frac{\sec(nx)\sin(nx)}{n}\, \mathrm{d}x\\ &= -\int \frac{\tan(nx)}{n}\, \mathrm{d}x\\ &= -\frac{\ln(\sec(nx))}{n^2} + C \end{aligned} \\ \begin{aligned} V_2(x) &= \int \frac{r(x)\cos(nx)}{W(x)}\, \mathrm{d}x \\ &= \int \frac{\sec(nx)\cos(nx)}{n}\, \mathrm{d}x\\ &= \int \frac{1}{n}\, \mathrm{d}x\\ &= \frac{x}{n} + C \end{aligned} \\ \textsf{The constant terms of the integration can be} \\\textsf{ignored since we are trying to find a non-constant} \\\textsf{solution to the differential equation} \\ \begin{aligned} \therefore y &= y_c + y_p \\ &=C_1\cos(nx) + C_2\sin(nx) - \frac{1}{n^2}\ln(\sec(nx))\cos(nx) +\frac{x}{n}\sin(nx) \end{aligned}


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