Question #97642
d^2y/d^2 = 1/x(x+1) + cosec^2x
1
Expert's answer
2019-10-31T10:56:18-0400

We need to solve the equation


y(x)=1x(x+1)+1sin2xy''(x) = \frac{1}{{x(x + 1)}} + \frac{1}{{{{\sin }^2}x}}

(we used that csc2x=1sin2x{\csc ^2}x = \frac{1}{{{{\sin }^2}x}})

To solve it, we need to integrate it twice. Let's do the first integration


y(x)=C1+dxx(x+1)+dxsin2xy'(x) = {C_1} + \int {\frac{{dx}}{{x(x + 1)}}} + \int {\frac{{dx}}{{{{\sin }^2}x}}}

To find the integral dxx(x+1)\int {\frac{{dx}}{{x(x + 1)}}} let's represent the fraction 1x(x+1)\frac{1}{{x(x + 1)}} as a sum of simple fractions 1x(x+1)=1x1x+1\frac{1}{{x(x + 1)}} = \frac{1}{x} - \frac{1}{{x + 1}} then


dxx(x+1)=dxxdxx+1=lnxlnx+1\int {\frac{{dx}}{{x(x + 1)}}} = \int {\frac{{dx}}{x}} - \int {\frac{{dx}}{{x + 1}}} = \ln \left| x \right| - \ln \left| {x + 1} \right|

The nex integral dxsin2x\int {\frac{{dx}}{{{{\sin }^2}x}}} can be found if we you notice that


dcot(x)dx=sinxsinxcosxsin2xcosx=11sin2xsin2x=1sin2x\frac{{d\cot (x)}}{{dx}} = \frac{{ - \sin x}}{{\sin x}} - \frac{{\cos x}}{{{{\sin }^2}x}}\cos x = - 1 - \frac{{1 - {{\sin }^2}x}}{{{{\sin }^2}x}} = - \frac{1}{{{{\sin }^2}x}}

It means that

dxsin2x=cotx\int {\frac{{dx}}{{{{\sin }^2}x}}} = - \cot x

We get

y(x)=C1+lnxlnx+1cotxy'(x) = {C_1} + \ln \left| x \right| - \ln \left| {x + 1} \right| - \cot x

Do the integration again


y(x)=C1x+C2+lnxdxlnx+1dxcotxdxy(x) = {C_1}x + {C_2} + \int {\ln \left| x \right|dx} - \int {\ln \left| {x + 1} \right|dx} - \int {\cot xdx}

First integral can be calculated by parts. Let x>0x > 0 we get


lnxdx=xlnxxdxx=x(lnx1)+c\int {\ln xdx} = x\ln x - \int {x \cdot \frac{{dx}}{x}} = x(\ln x - 1)+c

For x<0x < 0 using ξ=x\xi = - x we get


ln(x)dx=lnξdξ=ξ(lnξ1)+c=x(ln(x)1)+c\int {\ln ( - x)dx} = - \int {\ln \xi d\xi } = - \xi (\ln \xi - 1)+c = x(\ln ( - x) - 1)+c

Thus


lnxdx=x(lnx1)+c\int {\ln \left| x \right|dx} = x(\ln \left| x \right| - 1)+c

The second integral can be reduced to the first using η=x+1\eta = x + 1


lnx+1dx=lnηdη\int {\ln \left| {x + 1} \right|dx} = \int {\ln \left| \eta \right|d\eta }

Thus


lnx+1dx=(x+1)(lnx+11)+c\int {\ln \left| {x + 1} \right|dx} = (x + 1)(\ln \left| {x + 1} \right| - 1)+c

The last integral


cotxdx=cosxsinxdx=d(sinx)sinx=lnsinx+c\int {\cot xdx} = \int {\frac{{\cos x}}{{\sin x}}dx} = \int {\frac{{d(\sin x)}}{{\sin x}}} = \ln \left| {\sin x} \right|+c

And we get (we omit some terms because they can be included in C1x{C_1}x and C2{C_2})


y(x)=C1x+C2+xlnx(x+1)lnx+1lnsinxy(x) = {C_1}x + {C_2} + x\ln \left| x \right| - (x + 1)\ln \left| {x + 1} \right| - \ln \left| {\sin x} \right|





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