Solve the following second-order linear differential equation by the method of reduction of order. (D2+1)y=csc⁡ x

Name: Solve the following second-order linear differential equation by the method of reduction of order. (D2+1)y=csc⁡ x
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the following second-order linear differential equation by the method of reduction of order. (D2+1)y=csc⁡ x

Question

Solve the following second-order linear differential equation by the method of reduction of order.  (D2+1)y=csc⁡ x

jean2098 · Accepted Answer

Now the characteristic equation of the differential equation is:  m2+1=0  m=±i  Roots are complex. Therefore the complementary solution is:  yc=C1cos⁡(x)+C2sin⁡(x)  Now the fundamental solution set is:  y1=cos⁡(x) and y2=sin⁡(x)  Now the wronskian of the solution set is:  W(x)=y1y2y1y2  W(x)=1  Now using method of variation of parameter the particular solution to the given non−homogeneous differential equation is:  yp=−y1∫y2⋅csc⁡(x)W(x)dx+y2∫y1⋅csc⁡(x)W(x)dx  yp=−cos⁡(x)∫sin⁡(x)⋅csc⁡(x)dx+sin⁡(x)∫cos⁡(x)⋅csc⁡(x)dx  yp=−cos⁡(x)∫dx+sin⁡(x)∫cot⁡(x)dx  yp=−xcos⁡(x)+sin⁡(x)ln|sin⁡(x)|  Hence the general solution is:  y(x)=yc+yp  y(x)=C1cos⁡(x)+C2sin⁡(x)−xcos⁡(x)+sin⁡(x)ln|sin⁡(x)|

sirpsta3u · Accepted Answer

For the equation (D2+1)y=csc⁡(x) the function g(x)=sin⁡x is solution of the homogeneous part ,i.e. g″+g=0. Then take y=g(x)V(x) and substitute in the equation. Obtain (g″+g)V+2g′V′+gV″=1g. Then V″+2cot⁡(x)V′=1/(sin⁡x)2. This is a first order equation for V’(x). The integrating factor is (sin⁡x)2 and the solution is V′(x)=(1(sin⁡x)2) [ Integral of ((sin⁡x)2(sin⁡x)2)dx+C]=x+C(sin⁡x)2 Integrating obtain V(x)=ln⁡(sin⁡x)−(C+x)cot⁡(x)+D. Then the solution is y=sin⁡(x)[ln⁡(sin⁡x)+D]−(C+x)cos⁡(x).

karton · Accepted Answer

⇒D2y+y=cosecxThe auxiliary equation ism2+1=0⇒m=±iC.F.yc=c1cos⁡(x)+c2sin⁡(x)Let y1=cos⁡(x) and y2=sin⁡(x)cos2⁡(x)+sin2⁡(x)=1By variation of parameter method consider particular solution asyp=u(x)⋅y1+v(x)⋅y2where u(x)=∫−y2wcosec(x)dx=∫−sin⁡(x)1cosec(x)dx=−∫sin⁡(x)sin⁡(x)dx=−∫dx=−xand v(x)=∫y1wcosec(x)dx=∫cos⁡(x)1cosec(x)dx=∫cos⁡(x)sin⁡(x)dx=∫cos⁡(x)sin⁡(x)dx=∫cot⁡(x)dxln⁡(sin⁡x)Hence yp=u(x)⋅y1+v(x)⋅y2=−x⋅cos⁡(x)+ln⁡(sin⁡x)⋅(sin⁡x)The general solution isy=yc+yp=c1cos⁡(x)+c2sin⁡(x)−x.cos⁡(x)+sin⁡(x).ln⁡(sin⁡(x))

Solve the following second-order linear differential equation by the method

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