In [1]:

    

from sympy import *


    

In [2]:

    

init_printing()


    

In [3]:

    

x,k,c1,c2=symbols('x,k,c_1,c_2')
y=Function('y')(x)
x,k,c1,c2,y


    

    
Out[3]:





$$\left ( x, \quad k, \quad c_{1}, \quad c_{2}, \quad y{\left (x \right )}\right )$$

In [4]:

    

ecu6a=Eq(y,c1*x+c2*x**2)
ecu6a
z=ecu6a.rhs
z
ec1=Eq(y.diff(x),z.diff(x))
ec2=Eq(y.diff(x,2),z.diff(x,2))
subs1=solve([ec1,ec2],[c1,c2])
(ecu6a.subs(subs1)).simplify()


    

    
Out[4]:





$$y{\left (x \right )} = \frac{x}{2} \left(- x \frac{d^{2}}{d x^{2}}  y{\left (x \right )} + 2 \frac{d}{d x} y{\left (x \right )}\right)$$

In [5]:

    

ecu6b=Eq(y,c1*exp(k*x)+c2*exp(-k*x))
ecu6b


    

    
Out[5]:





$$y{\left (x \right )} = c_{1} e^{k x} + c_{2} e^{- k x}$$

In [6]:

    

z=c1*exp(k*x)+c2*exp(-k*x)
z


    

    
Out[6]:





$$c_{1} e^{k x} + c_{2} e^{- k x}$$

In [7]:

    

z.diff(x)


    

    
Out[7]:





$$c_{1} k e^{k x} - c_{2} k e^{- k x}$$

In [8]:

    

ec1=Eq(y.diff(x),z.diff(x))
ec2=Eq(y.diff(x,2),z.diff(x,2))
ec1, ec2


    

    
Out[8]:





$$\left ( \frac{d}{d x} y{\left (x \right )} = c_{1} k e^{k x} - c_{2} k e^{- k x}, \quad \frac{d^{2}}{d x^{2}}  y{\left (x \right )} = k^{2} \left(c_{1} e^{k x} + c_{2} e^{- k x}\right)\right )$$

In [9]:

    

subs1=solve([ec1,ec2],[c1,c2])
subs1


    

    
Out[9]:





$$\left \{ c_{1} : \frac{e^{- k x}}{2 k^{2}} \left(k \frac{d}{d x} y{\left (x \right )} + \frac{d^{2}}{d x^{2}}  y{\left (x \right )}\right), \quad c_{2} : \frac{e^{k x}}{2 k^{2}} \left(- k \frac{d}{d x} y{\left (x \right )} + \frac{d^{2}}{d x^{2}}  y{\left (x \right )}\right)\right \}$$

In [10]:

    

(ecu6b.subs(subs1)).simplify()


    

    
Out[10]:





$$y{\left (x \right )} = \frac{1}{k^{2}} \frac{d^{2}}{d x^{2}}  y{\left (x \right )}$$

In [11]:

    

ecu6c=Eq(y,c1*sin(k*x)+c2*cos(-k*x))
ecu6c


    

    
Out[11]:





$$y{\left (x \right )} = c_{1} \sin{\left (k x \right )} + c_{2} \cos{\left (k x \right )}$$

In [12]:

    

z=ecu6c.rhs
z


    

    
Out[12]:





$$c_{1} \sin{\left (k x \right )} + c_{2} \cos{\left (k x \right )}$$

In [13]:

    

ec1=Eq(y.diff(x),z.diff(x))
ec2=Eq(y.diff(x,2),z.diff(x,2))
ec1, ec2


    

    
Out[13]:





$$\left ( \frac{d}{d x} y{\left (x \right )} = c_{1} k \cos{\left (k x \right )} - c_{2} k \sin{\left (k x \right )}, \quad \frac{d^{2}}{d x^{2}}  y{\left (x \right )} = - k^{2} \left(c_{1} \sin{\left (k x \right )} + c_{2} \cos{\left (k x \right )}\right)\right )$$

In [14]:

    

subs1=solve([ec1,ec2],[c1,c2])
subs1


    

    
Out[14]:





$$\left \{ c_{1} : \frac{1}{k^{2}} \left(k \cos{\left (k x \right )} \frac{d}{d x} y{\left (x \right )} - \sin{\left (k x \right )} \frac{d^{2}}{d x^{2}}  y{\left (x \right )}\right), \quad c_{2} : - \frac{1}{k^{2}} \left(k \sin{\left (k x \right )} \frac{d}{d x} y{\left (x \right )} + \cos{\left (k x \right )} \frac{d^{2}}{d x^{2}}  y{\left (x \right )}\right)\right \}$$

In [15]:

    

(ecu6c.subs(subs1)).simplify()


    

    
Out[15]:





$$y{\left (x \right )} = - \frac{1}{k^{2}} \frac{d^{2}}{d x^{2}}  y{\left (x \right )}$$

In [16]:

    

z1=x
z2=x**2
((z1.diff(x,2)-(2/x)*z1.diff(x)+(2/(x**2))*z1)).simplify(), ((z2.diff(x,2)-(2/x)*z2.diff(x)+(2/(x**2))*z2)).simplify()


    

    
Out[16]:





$$\left ( 0, \quad 0\right )$$

In [17]:

    

solve([Eq(3,c1+c2), Eq(5,c1+2*c2)],[c1, c2])


    

    
Out[17]:





$$\left \{ c_{1} : 1, \quad c_{2} : 2\right \}$$

In [18]:

    

v=Function('v')(x)
y2=v*x
y2, (1-x**2)*y.diff(x,2)-2*x*y.diff(x)+2*y


    

    
Out[18]:





$$\left ( x v{\left (x \right )}, \quad - 2 x \frac{d}{d x} y{\left (x \right )} + \left(- x^{2} + 1\right) \frac{d^{2}}{d x^{2}}  y{\left (x \right )} + 2 y{\left (x \right )}\right )$$

In [19]:

    

ecv=((1-x**2)*y2.diff(x,2)-2*x*y2.diff(x)+2*y2).simplify()
ecv


    

    
Out[19]:





$$- x^{3} \frac{d^{2}}{d x^{2}}  v{\left (x \right )} - 4 x^{2} \frac{d}{d x} v{\left (x \right )} + x \frac{d^{2}}{d x^{2}}  v{\left (x \right )} + 2 \frac{d}{d x} v{\left (x \right )}$$

In [20]:

    

u=Function('u')(x)
ecu=Eq(u.diff(x)+2*(1/x-x/(1-x**2))*u)
ecu


    

    
Out[20]:





$$\left(- \frac{2 x}{- x^{2} + 1} + \frac{2}{x}\right) u{\left (x \right )} + \frac{d}{d x} u{\left (x \right )} = 0$$

In [21]:

    

dsolve(ecu,u)


    

    
Out[21]:





$$u{\left (x \right )} = \frac{C_{1}}{x^{2} \left(x^{2} - 1\right)}$$

In [22]:

    

V=integrate(dsolve(ecu,u).rhs,x)
V


    

    
Out[22]:





$$C_{1} \left(\frac{1}{2} \log{\left (x - 1 \right )} - \frac{1}{2} \log{\left (x + 1 \right )} + \frac{1}{x}\right)$$

In [23]:

    

#Luego y_3=V*x resuelve (1-x^2)y''-2xy'+2y=0
y3=V*x
y3.simplify(), ((1-x**2)*y3.diff(x,2)-2*x*y3.diff(x)+2*y3).simplify()


    

    
Out[23]:





$$\left ( \frac{C_{1}}{2} \left(x \left(\log{\left (x - 1 \right )} - \log{\left (x + 1 \right )}\right) + 2\right), \quad 0\right )$$

In [53]:

    

#verifiquemos que 1/\sqrt(x) es solucion de x^2y''+xy'+(x^2-1/4)y=0
x=symbols('x')
y1=(x)**(-Rational(1,2))*sin(x)
yy=(x**2*(y1.diff(x,2))+x*y1.diff(x)+(x**2-Rational(1,4))*y1).simplify()
y1, yy


    

    
Out[53]:





$$\left ( \frac{1}{\sqrt{x}} \sin{\left (x \right )}, \quad 0\right )$$

In [56]:

    

v=Function('v')(x)
y2=v*y1
ecc=(x**2*(y2.diff(x,2))+x*y2.diff(x)+(x**2-Rational(1,4))*y2).simplify()
y2, ecc


    

    
Out[56]:





$$\left ( \frac{1}{\sqrt{x}} v{\left (x \right )} \sin{\left (x \right )}, \quad x^{\frac{3}{2}} \left(\sin{\left (x \right )} \frac{d^{2}}{d x^{2}}  v{\left (x \right )} + 2 \cos{\left (x \right )} \frac{d}{d x} v{\left (x \right )}\right)\right )$$

In [58]:

    

u=Function('u')(x)
ecu=Eq(sin(x)*u.diff(x)+2*cos(x)*u)
ecu


    

    
Out[58]:





$$2 u{\left (x \right )} \cos{\left (x \right )} + \sin{\left (x \right )} \frac{d}{d x} u{\left (x \right )} = 0$$

In [62]:

    

sol=dsolve(ecu,u)
sol


    

    
Out[62]:





$$u{\left (x \right )} = \frac{C_{1}}{\sin^{2}{\left (x \right )}}$$

In [64]:

    

integrate(sol.rhs,x)


    

    
Out[64]:





$$- \frac{C_{1} \cos{\left (x \right )}}{\sin{\left (x \right )}}$$

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In [45]:

    

f=Function('f')(x)
v=Function('v')(x)
y2=v*x
y2


    

    
Out[45]:





$$\left(C_{1} \cos{\left (\sqrt{C} t \right )} + C_{2} \sin{\left (\sqrt{C} t \right )} + L - \frac{g}{C}\right) v{\left (C_{1} \cos{\left (\sqrt{C} t \right )} + C_{2} \sin{\left (\sqrt{C} t \right )} + L - \frac{g}{C} \right )}$$

In [26]:

    

(y2.diff(x,2)-x*f*y2.diff(x)+f*y2).simplify()


    

    
Out[26]:





$$- x^{2} f{\left (x \right )} \frac{d}{d x} v{\left (x \right )} + x \frac{d^{2}}{d x^{2}}  v{\left (x \right )} + 2 \frac{d}{d x} v{\left (x \right )}$$

In [27]:

    

u=Function('u')(x)


    

In [28]:

    

c1,c2=symbols('c_1,c_2')
solve([Eq(2,c1*exp(1)+c2*exp(-9)),Eq(0,c1*exp(1)-9*c2*exp(-9))],[c1,c2])


    

    
Out[28]:





$$\left \{ c_{1} : \frac{9}{5 e}, \quad c_{2} : \frac{e^{9}}{5}\right \}$$

In [29]:

    

c1,c2=symbols('c_1,c_2')
z=1/x*(c1*cos(3*log(x))+c2*sin(3*log(x)))
z


    

    
Out[29]:





$$\frac{1}{x} \left(c_{1} \cos{\left (3 \log{\left (x \right )} \right )} + c_{2} \sin{\left (3 \log{\left (x \right )} \right )}\right)$$

In [30]:

    

(x**2*z.diff(x,2)+3*x*z.diff(x)+10*z).simplify()


    

    
Out[30]:





$$0$$

In [31]:

    

c1,c2=symbols('c_1,c_2')
z=1/(x**2)*(c1*log(x)+c2)
z


    

    
Out[31]:





$$\frac{1}{x^{2}} \left(c_{1} \log{\left (x \right )} + c_{2}\right)$$

In [32]:

    

(2*x**2*z.diff(x,2)+10*x*z.diff(x)+8*z).simplify()


    

    
Out[32]:





$$0$$

In [33]:

    

b,k=symbols('b,k')
z=1/(k**2-b**2)*sin(b*x)
z


    

    
Out[33]:





$$\frac{\sin{\left (b x \right )}}{- b^{2} + k^{2}}$$

In [34]:

    

((z.diff(x,2)+k**2*z)-sin(b*x)).simplify()


    

    
Out[34]:





$$0$$

In [35]:

    

integrate(x*exp(-x),x), integrate(x**2*exp(-x),x)


    

    
Out[35]:





$$\left ( \left(- x - 1\right) e^{- x}, \quad \left(- x^{2} - 2 x - 2\right) e^{- x}\right )$$

In [36]:

    

C1=-integrate(Rational(1,2)*tan(2*x)*sin(2*x),x).simplify()
C2=integrate(Rational(1,2)*tan(2*x)*cos(2*x),x).simplify()
C1, C2


    

    
Out[36]:





$$\left ( \frac{1}{8} \log{\left (\sin{\left (2 x \right )} - 1 \right )} - \frac{1}{8} \log{\left (\sin{\left (2 x \right )} + 1 \right )} + \frac{1}{4} \sin{\left (2 x \right )}, \quad - \frac{1}{4} \cos{\left (2 x \right )}\right )$$

In [37]:

    

Y=(C1*cos(2*x)+C2*sin(2*x)).simplify()
Y


    

    
Out[37]:





$$\frac{1}{8} \left(\log{\left (\sin{\left (2 x \right )} - 1 \right )} - \log{\left (\sin{\left (2 x \right )} + 1 \right )}\right) \cos{\left (2 x \right )}$$

In [38]:

    

(Y.diff(x,2)+4*Y).simplify()


    

    
Out[38]:





$$\tan{\left (2 x \right )}$$

In [39]:

    

m,g,L,r,rhoa,C=symbols('m,g,L,r,rho_a,C',positive=true)
C1,C2,t=symbols('C_1,C_2,t')
x=Function('x')(t)


    

In [40]:

    

x=C1*cos(sqrt(C)*t)+C2*sin(sqrt(C)*t)-g/C+L
x


    

    
Out[40]:





$$C_{1} \cos{\left (\sqrt{C} t \right )} + C_{2} \sin{\left (\sqrt{C} t \right )} + L - \frac{g}{C}$$

In [41]:

    

ec1=Eq(x.subs(t,0),0)
ec2=Eq(x.diff(t).subs(t,0),0)
ec1, ec2


    

    
Out[41]:





$$\left ( C_{1} + L - \frac{g}{C} = 0, \quad \sqrt{C} C_{2} = 0\right )$$