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y'' + 3y = 5lnx help?

Guest Nov 25, 2015

Best Answer 

 #1
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+15

y'' + 3y = 5lnx help?

This is a second-order linear ordinary differential equation. The solution to it is quite involved.

 

Solve ( d^2 y(x))/( dx^2)+3 y(x) = 5 log(x):
The general solution will be the sum of the complementary solution and particular solution.
Find the complementary solution by solving ( d^2 y(x))/( dx^2)+3 y(x)  =  0:
Assume a solution will be proportional to e^(lambda x) for some constant lambda.
Substitute y(x)  =  e^(lambda x) into the differential equation:
( d^2 )/( dx^2)(e^(lambda x))+3 e^(lambda x)  =  0
Substitute ( d^2 )/( dx^2)(e^(lambda x))  =  lambda^2 e^(lambda x):
lambda^2 e^(lambda x)+3 e^(lambda x)  =  0
Factor out e^(lambda x):
(lambda^2+3) e^(lambda x)  =  0
Since e^(lambda x) !=0 for any finite lambda, the zeros must come from the polynomial:
lambda^2+3  =  0
Solve for lambda:
lambda = i sqrt(3) or lambda = -i sqrt(3)
The roots lambda  =  ± i sqrt(3) give y_1(x) = c_1 e^(i sqrt(3) x), y_2(x) = c_2 e^(-i sqrt(3) x) as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(x)  =  y_1(x)+y_2(x)  =  c_1 e^(i sqrt(3) x)+c_2/e^(i sqrt(3) x)
Apply Euler's identity e^(alpha+i beta) = e^alpha cos(beta)+i e^alpha sin(beta):
y(x)  =  c_1 (cos(sqrt(3) x)+i sin(sqrt(3) x))+c_2 (cos(sqrt(3) x)-i sin(sqrt(3) x))
Regroup terms:
y(x)  =  (c_1+c_2) cos(sqrt(3) x)+i (c_1-c_2) sin(sqrt(3) x)
Redefine c_1+c_2 as c_1 and i (c_1-c_2) as c_2, since these are arbitrary constants:
y(x)  =  c_1 cos(sqrt(3) x)+c_2 sin(sqrt(3) x)
Determine the particular solution to ( d^2 y(x))/( dx^2)+3 y(x)  =  5 log(x) by variation of parameters:
List the basis solutions in y_c(x):
y_(b_1)(x) = cos(sqrt(3) x) and y_(b_2)(x) = sin(sqrt(3) x)
Compute the Wronskian of y_(b_1)(x) and y_(b_2)(x):
(script capital w)(x)  =  |cos(sqrt(3) x) | sin(sqrt(3) x)
( d)/( dx)(cos(sqrt(3) x)) | ( d)/( dx)(sin(sqrt(3) x))|  =  |cos(sqrt(3) x) | sin(sqrt(3) x)
-(sqrt(3) sin(sqrt(3) x)) | sqrt(3) cos(sqrt(3) x)|  =  sqrt(3)
Let f(x) = 5 log(x):
Let v_1(x) = - integral (f(x) y_(b_2)(x))/((script capital w)(x)) dx and v_2(x) =  integral (f(x) y_(b_1)(x))/((script capital w)(x)) dx:
The particular solution will be given by:
y_p(x)  =  v_1(x) y_(b_1)(x)+v_2(x) y_(b_2)(x)
Compute v_1(x):
v_1(x)  =  - integral (5 log(x) sin(sqrt(3) x))/sqrt(3) dx  =  -5/3 (Ci(sqrt(3) x)-cos(sqrt(3) x) log(x))
Compute v_2(x):
v_2(x)  =   integral (5 cos(sqrt(3) x) log(x))/sqrt(3) dx  =  5/3 (log(x) sin(sqrt(3) x)-Si(sqrt(3) x))
The particular solution is thus:
y_p(x)  =  v_1(x) y_(b_1)(x)+v_2(x) y_(b_2)(x)  =  -5/3 cos(sqrt(3) x) (Ci(sqrt(3) x)-cos(sqrt(3) x) log(x))+5/3 (log(x) sin(sqrt(3) x)-Si(sqrt(3) x)) sin(sqrt(3) x)
Simplify:
y_p(x)  =  -5/3 (cos(sqrt(3) x) Ci(sqrt(3) x)-log(x)+sin(sqrt(3) x) Si(sqrt(3) x))
The general solution is given by:
Answer: | 
| y(x)  =  y_c(x) + y_p(x)  =  c_1 cos(sqrt(3) x)+c_2 sin(sqrt(3) x)-5/3 (cos(sqrt(3) x) Ci(sqrt(3) x)-log(x)+sin(sqrt(3) x) Si(sqrt(3) x))

Guest Nov 25, 2015
 #1
avatar
+15
Best Answer

y'' + 3y = 5lnx help?

This is a second-order linear ordinary differential equation. The solution to it is quite involved.

 

Solve ( d^2 y(x))/( dx^2)+3 y(x) = 5 log(x):
The general solution will be the sum of the complementary solution and particular solution.
Find the complementary solution by solving ( d^2 y(x))/( dx^2)+3 y(x)  =  0:
Assume a solution will be proportional to e^(lambda x) for some constant lambda.
Substitute y(x)  =  e^(lambda x) into the differential equation:
( d^2 )/( dx^2)(e^(lambda x))+3 e^(lambda x)  =  0
Substitute ( d^2 )/( dx^2)(e^(lambda x))  =  lambda^2 e^(lambda x):
lambda^2 e^(lambda x)+3 e^(lambda x)  =  0
Factor out e^(lambda x):
(lambda^2+3) e^(lambda x)  =  0
Since e^(lambda x) !=0 for any finite lambda, the zeros must come from the polynomial:
lambda^2+3  =  0
Solve for lambda:
lambda = i sqrt(3) or lambda = -i sqrt(3)
The roots lambda  =  ± i sqrt(3) give y_1(x) = c_1 e^(i sqrt(3) x), y_2(x) = c_2 e^(-i sqrt(3) x) as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(x)  =  y_1(x)+y_2(x)  =  c_1 e^(i sqrt(3) x)+c_2/e^(i sqrt(3) x)
Apply Euler's identity e^(alpha+i beta) = e^alpha cos(beta)+i e^alpha sin(beta):
y(x)  =  c_1 (cos(sqrt(3) x)+i sin(sqrt(3) x))+c_2 (cos(sqrt(3) x)-i sin(sqrt(3) x))
Regroup terms:
y(x)  =  (c_1+c_2) cos(sqrt(3) x)+i (c_1-c_2) sin(sqrt(3) x)
Redefine c_1+c_2 as c_1 and i (c_1-c_2) as c_2, since these are arbitrary constants:
y(x)  =  c_1 cos(sqrt(3) x)+c_2 sin(sqrt(3) x)
Determine the particular solution to ( d^2 y(x))/( dx^2)+3 y(x)  =  5 log(x) by variation of parameters:
List the basis solutions in y_c(x):
y_(b_1)(x) = cos(sqrt(3) x) and y_(b_2)(x) = sin(sqrt(3) x)
Compute the Wronskian of y_(b_1)(x) and y_(b_2)(x):
(script capital w)(x)  =  |cos(sqrt(3) x) | sin(sqrt(3) x)
( d)/( dx)(cos(sqrt(3) x)) | ( d)/( dx)(sin(sqrt(3) x))|  =  |cos(sqrt(3) x) | sin(sqrt(3) x)
-(sqrt(3) sin(sqrt(3) x)) | sqrt(3) cos(sqrt(3) x)|  =  sqrt(3)
Let f(x) = 5 log(x):
Let v_1(x) = - integral (f(x) y_(b_2)(x))/((script capital w)(x)) dx and v_2(x) =  integral (f(x) y_(b_1)(x))/((script capital w)(x)) dx:
The particular solution will be given by:
y_p(x)  =  v_1(x) y_(b_1)(x)+v_2(x) y_(b_2)(x)
Compute v_1(x):
v_1(x)  =  - integral (5 log(x) sin(sqrt(3) x))/sqrt(3) dx  =  -5/3 (Ci(sqrt(3) x)-cos(sqrt(3) x) log(x))
Compute v_2(x):
v_2(x)  =   integral (5 cos(sqrt(3) x) log(x))/sqrt(3) dx  =  5/3 (log(x) sin(sqrt(3) x)-Si(sqrt(3) x))
The particular solution is thus:
y_p(x)  =  v_1(x) y_(b_1)(x)+v_2(x) y_(b_2)(x)  =  -5/3 cos(sqrt(3) x) (Ci(sqrt(3) x)-cos(sqrt(3) x) log(x))+5/3 (log(x) sin(sqrt(3) x)-Si(sqrt(3) x)) sin(sqrt(3) x)
Simplify:
y_p(x)  =  -5/3 (cos(sqrt(3) x) Ci(sqrt(3) x)-log(x)+sin(sqrt(3) x) Si(sqrt(3) x))
The general solution is given by:
Answer: | 
| y(x)  =  y_c(x) + y_p(x)  =  c_1 cos(sqrt(3) x)+c_2 sin(sqrt(3) x)-5/3 (cos(sqrt(3) x) Ci(sqrt(3) x)-log(x)+sin(sqrt(3) x) Si(sqrt(3) x))

Guest Nov 25, 2015

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