Differential equations(x+1)2y″+(x+1)y′+y=x2+2sin⁡(ln⁡(x+1)),y(0)=15,y′(0)=2

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Demetrius Kaufman · Accepted Answer

Step 1y″+y′(x+1)+y(x+1)2=x2+2sin⁡(ln⁡(x+1))(x+1)2First of all , I found the equation solution of y″+y′(x+1)+y(x+1)2=0y=c1cos⁡(ln⁡(x+1))+c2sin⁡(ln⁡(x+1))I try to solve this ode using the variation of parameters theoremGet this system equation:1) c1′cos⁡(ln⁡(x+1))+c2′sin⁡(ln⁡(x+1))=02) −c1′sin⁡(ln⁡(x+1))+c2′cos⁡(ln⁡(x+1))=x2+2sin⁡(ln⁡(x+1))Multiply 1) by sin⁡(ln⁡(x+1)), 2) by cos⁡(ln⁡(x+1))x+1.Step 2By addtion i get:c2′=cos⁡(ln⁡(x+1))[x2+2sin⁡(ln⁡(x+1))](x+1)I do not know how I get c2 by an integral ?

undodaonePvopxl24 · Accepted Answer

Step 1The two equations you have using variation of parameters is incorrect. Given a second-order linear inhomogeneous DE&amp;nbsp;y&amp;nbsp;″+p(x)y′+q(x)y=f(x)&amp;nbsp;with homogeneous solution&amp;nbsp;yc(x)=c1y1(x)+c2y2(x), a particular solution&amp;nbsp;yp&amp;nbsp;is given by&amp;nbsp;yp(x)=u1(x)y1(x)+u2(x)y2(x), where&amp;nbsp;u1&amp;nbsp;and&amp;nbsp;u2&amp;nbsp;satisfy                                        u            1            ′                                y            1                    +                      u            2            ′                                y            2                                              =          0                                                  u            1            ′                                y            1            ′                    +                      u            2            ′                                y            2            ′                                              =          f          (          x          )          .                    Let&amp;nbsp;y1=cos(ln(x+1))&amp;nbsp;&amp;nbsp;and&amp;nbsp;&amp;nbsp;y2=sin(ln(x+1)).We have thatStep 2Multiplying (1) by&amp;nbsp;sin⁡(ln⁡(x+1)), (2) by&amp;nbsp;(x+1)cos⁡(ln⁡(x+1)), and adding the two resulting equations, we getu2′&amp;nbsp;=x2+2sin⁡(ln⁡(x+1))x+1⋅cos⁡(ln⁡(x+1)).Step 3Using the substitution&amp;nbsp;z=ln⁡(x+1), we see thatu2=∫[(ez−1)2+2sin⁡z]cos⁡z,&amp;nbsp;dz&amp;nbsp;.Computing this integral is a standard exercise now. Can you take it from here?

Differential equations \(\displaystyle{\left({x}+{1}\right)}^{{2}}{y}{''}+{\left({x}+{1}\right)}{y}'+{y}={x}^{{2}}+{2}{\sin{{\left({\ln{{\left({x}+{1}\right)}}}\right)}}},{y}{\left({0}\right)}={\frac{{{1}}}{{{5}}}},{y}'{\left({0}\right)}={2}\)

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