Question

Solve x2 d2y/dx2+7x dy/dx + 13y = logx​

Answer 1

Concept: It is a non-homogenous linear differential equation. The answer will be a complementary solution + a particular integral solution.

Given:

$x^{2} (d^{2}y/dx^{2} ) +7x (dy/dx) +13y=logx$

Answer:

Since the given equation is a non-homogenous linear differential equation,

put $x=e^{z}$

     $z=logx$

     $\frac{dz}{dx} =\frac{1}{x}$

and we know, we can write:

$x \frac{d}{dx} =D$

and,$x^{2} \frac{d^{2} }{dx^{2} } = D(D-1)$

then, the equation will become,

$D(D-1)y +7Dy+13y=z$

Then, the Auxiliary equation will be:

$m^{2}+6m+13=0$

then,

$m=-3-i2,-3+i2$

∴$C.F.= e^{-3x} (A cos2x+B sin2x)$

$P.F. =\frac{z}{D^{2}+6D+13 } \\ \\ =\frac{1}{13} (1+(\frac{D^{2} +6D}{13}))^{-1} z$

$=\frac{1}{13} (1-(\frac{D^{2} +6D}{13}) +....)} z\\\\=\frac{1}{13}(z-\frac{6}{13} )\\=\frac{z}{13} -\frac{6}{169}$

then,

$y=e^{-3x} (A cos2x+B sin2x)+\frac{z}{13} -\frac{6}{169}$

Hence, the correct answer to the given differential equation will be $y=e^{-3x} (A cos2x+B sin2x)+\frac{z}{13} -\frac{6}{169}$.

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Answer 2

Answer:

e^-7/2x( C1cos√3 + C2sin√3x) + 1/ D² +7D +13 * (logx)

Step-by-step explanation:

Given :  x² d²y/dx²+7x dy/dx + 13y = logx​ ----------- equation 1

⇒ this is homogeneous differential equation

⇒ so put x = e^z ; taking (log base e ) both side we get

⇒ z = log x ; differentiating both side ---------- equation 2

⇒ dz/dx = 1/x

⇒ from equation 1 , taking y as common

⇒ y[ x²d²/dx² + 7xd/dx + 13 ] = log x

⇒ let xd/dx = D  ,

⇒ y[ D² + 7D +13] = z

⇒ The auxiliary equation is

D² +7D +13 = 0 ; Let D = m

: m² + 7m + 13 = 0

;  on solving using quadratic formula i.e  -b ±√b²- 4ac÷2a

we get ; m =  -7 + √3i /2  ,  -7-√3i /2

C.F for imaginary solution is ; e^-7/2x( C1cos√3 + C2sin√3x)

now for P.I = 1/f(D)  * Q(x)  = 0

NOW = 1/ D² +7D +13 * (z)

→  final solution / complete solution = CF + PI

= e^-7/2x( C1cos√3 + C2sin√3x) + 1/ D² +7D +13 * (logx)

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