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Cauchy's homogeneous linear differential form is given as x2d2y/dx2 - 3xdy/dx + 5y = x2 sin (log x).
If z = log x, then the particular integral will be is
  • a)
    1/2e-2zzcos z
  • b)
    1/2e2zcos z
  • c)
    -1/2e2zzcos z
  • d)
    -1/2e-2zcos z
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Cauchy's homogeneous linear differential form is given as x2d2y/dx2 -...
Solution:

Given differential equation is x^2(d^2y/dx^2) - 3x(dy/dx) + 5y = x^2 sin(logx)

Let z = logx, then x = e^z and dx/dz = e^z

Using chain rule, we get dy/dz = (dy/dx) * (dx/dz)

Differentiating w.r.t z again, we get d^2y/dz^2 = d/dz[(dy/dx) * (dx/dz)] = [(d^2y/dx^2) * (dx/dz)^2] + [(dy/dx) * d^2x/dz^2]

Substituting x = e^z and dx/dz = e^z, we get d^2y/dz^2 = e^2z(d^2y/dx^2) + e^z(dy/dx)

Substituting these values in the given differential equation, we get

e^2z(d^2y/dx^2) - 3e^z(dy/dx) + 5y = e^2z sin(z)

Dividing both sides by e^2z, we get

(d^2y/dx^2) - 3(dy/dx)/x + 5y/x^2 = sin(z)

This is a homogeneous linear differential equation of second order with constant coefficients.

Particular Integral:

Since the given equation is a homogeneous linear differential equation with constant coefficients, we can use the method of undetermined coefficients to find the particular integral.

Let the particular integral be of the form y_p = A e^(-2z) cos(z) + B e^(-2z) sin(z)

Differentiating w.r.t z, we get

dy_p/dz = (-2A e^(-2z) + B e^(-2z)) cos(z) + (-A e^(-2z) - 2B e^(-2z)) sin(z)

Differentiating again w.r.t z, we get

d^2y_p/dz^2 = (4A e^(-2z) - 4B e^(-2z)) cos(z) + (-4A e^(-2z) - 4B e^(-2z)) sin(z)

Substituting these values in the given differential equation, we get

(4A e^(-2z) - 4B e^(-2z)) cos(z) + (-4A e^(-2z) - 4B e^(-2z)) sin(z) - 3[(-2A e^(-2z) + B e^(-2z)) cos(z) + (-A e^(-2z) - 2B e^(-2z)) sin(z)]/e^z + 5[A e^(-2z) cos(z) + B e^(-2z) sin(z)]/e^(2z) = sin(z)

Simplifying the above equation, we get

(-4A + 3B) cos(z) + (-4B - 3A) sin(z) = sin(z)

Comparing coefficients of cos(z) and sin(z), we get

-4A + 3B = 0 and -4B - 3
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Community Answer
Cauchy's homogeneous linear differential form is given as x2d2y/dx2 -...
Put z= log x and D = d/dz
Then equation becomes
D (D - 1)y - 3Dy + 5y = e2z sin z
⇒ (D2 - 4D + 5)y = e2zsin z
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Cauchy's homogeneous linear differential form is given as x2d2y/dx2 - 3xdy/dx + 5y = x2 sin (log x).If z = log x, then the particular integral will be isa)1/2e-2zzcos zb)1/2e2zcos zc)-1/2e2zzcos zd)-1/2e-2zcos zCorrect answer is option 'C'. Can you explain this answer?
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