Solve `x^2 (d^2y)/dx^2+3x dy/dx+3y =(log x.cos (log x))/x`
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Solution
Given that,
`x^2( d^2y)/dx^2+3x dy/dx+3y=(log x.cos (log x))/x`
Putting z = log x and x = ez, we get
`[D(D-1)+3D+3]y=e^-z.z. cos z`
`[D^2+2D+3]y=e^-z.z.cos z`
∴ ` "The A.E is" D^2+2D+3=0`
∴ `D= (-2+-2sqrt2.i)/2=-1+- sqrt2.i`
∴`" The C.F. is y"= e^-z.z. cos z`
=`e^-z 1/((D-1)^2+2(D-1)+3).z.cos z= e^-z. 1/(D^2+2).z.cos z`
= `e^-z [Z-1/(D^2+2).2D].1/(D^2+2). cos z`
=`e_z [z-1/(D^2+2).2D]cos z=e^-z [z cosz+1/(D^2+2).2 sin z]`
=` e^-z[z cos z+2 sin z]`
The complete solution is,
`y=C.F.+P.I.`
`y=e^-z (C_1 cos sqrt2z+C_2 sin sqrt2z)+e^-z[z cos z+2 sin z]`
`Y=1/x(C_1 cos sqrt2 log x+C_2 sin sqrt2 log)+1/x [log x cos log x+2 sin log x]`
Concept: Equations Reducible to Exact Form by Using Integrating Factors
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