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The general solution of (x2 D2 – xD), y= 0 is :
- a)y = C1 + C2 ex
- b)y = C1 + C2 x2
- c)y = C1 x + C2 x2
- d)y = C1 + C2 x
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The general solution of(x2D2– xD), y=0 is :a)y = C1+ C2exb)y = C...
The correct answer is: y = C1 + C2 x
Most Upvoted Answer
The general solution of(x2D2– xD), y=0 is :a)y = C1+ C2exb)y = C...
+ 4xD + 3y = 0) can be found by assuming a solution of the form y = e^(mx).
Substituting this into the differential equation gives:
x^2(m^2)e^(mx) + 4xme^(mx) + 3e^(mx) = 0
Dividing through by e^(mx) gives:
m^2x^2 + 4mx + 3 = 0
This is a quadratic equation in m, which can be solved using the quadratic formula:
m = (-4x ± √(16x^2 - 4(3)(x^2))) / (2x^2)
Simplifying this gives:
m = (-2 ± √1) / x
m1 = -1/x, m2 = -3/x
Therefore, the general solution is a linear combination of the two solutions:
y = c1e^(-x) + c2e^(-3x)
where c1 and c2 are arbitrary constants determined by initial or boundary conditions.
Substituting this into the differential equation gives:
x^2(m^2)e^(mx) + 4xme^(mx) + 3e^(mx) = 0
Dividing through by e^(mx) gives:
m^2x^2 + 4mx + 3 = 0
This is a quadratic equation in m, which can be solved using the quadratic formula:
m = (-4x ± √(16x^2 - 4(3)(x^2))) / (2x^2)
Simplifying this gives:
m = (-2 ± √1) / x
m1 = -1/x, m2 = -3/x
Therefore, the general solution is a linear combination of the two solutions:
y = c1e^(-x) + c2e^(-3x)
where c1 and c2 are arbitrary constants determined by initial or boundary conditions.
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Community Answer
The general solution of(x2D2– xD), y=0 is :a)y = C1+ C2exb)y = C...
Use Cauchy Euler method. In this we put x=e^z.
We get xdy/dx=dz/dx
Similarly, x^2 d2y/dz2=d2z/dx2
& Then proceed
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The general solution of(x2D2– xD), y=0 is :a)y = C1+ C2exb)y = C1+ C2x2c)y = C1x +C2x2d)y = C1+ C2xCorrect answer is option 'C'. Can you explain this answer?
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