We need to solve the equation
y′′(x)=x(x+1)1+sin2x1 (we used that csc2x=sin2x1)
To solve it, we need to integrate it twice. Let's do the first integration
y′(x)=C1+∫x(x+1)dx+∫sin2xdx To find the integral ∫x(x+1)dx let's represent the fraction x(x+1)1 as a sum of simple fractions x(x+1)1=x1−x+11 then
∫x(x+1)dx=∫xdx−∫x+1dx=ln∣x∣−ln∣x+1∣ The nex integral ∫sin2xdx can be found if we you notice that
dxdcot(x)=sinx−sinx−sin2xcosxcosx=−1−sin2x1−sin2x=−sin2x1 It means that
∫sin2xdx=−cotx We get
y′(x)=C1+ln∣x∣−ln∣x+1∣−cotx Do the integration again
y(x)=C1x+C2+∫ln∣x∣dx−∫ln∣x+1∣dx−∫cotxdx First integral can be calculated by parts. Let x>0 we get
∫lnxdx=xlnx−∫x⋅xdx=x(lnx−1)+c For x<0 using ξ=−x we get
∫ln(−x)dx=−∫lnξdξ=−ξ(lnξ−1)+c=x(ln(−x)−1)+c Thus
∫ln∣x∣dx=x(ln∣x∣−1)+c The second integral can be reduced to the first using η=x+1
∫ln∣x+1∣dx=∫ln∣η∣dη Thus
∫ln∣x+1∣dx=(x+1)(ln∣x+1∣−1)+c The last integral
∫cotxdx=∫sinxcosxdx=∫sinxd(sinx)=ln∣sinx∣+c And we get (we omit some terms because they can be included in C1x and C2)
y(x)=C1x+C2+xln∣x∣−(x+1)ln∣x+1∣−ln∣sinx∣
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