Ecuaciones

Solve the linear equation
-3 y(x)+( dy(x))/( dx) = e^(2 x),

such that y(0) = 0:
Let mu(x) = exp( integral -3 dx) = e^(-3 x).
Multiply both sides by mu(x):
(-3 e^(-3 x)) y(x)+e^(-3 x) ( dy(x))/( dx) = e^(-x)
Substitute -3 e^(-3 x) = ( d)/( dx)(e^(-3 x)):
y(x) ( d)/( dx)(e^(-3 x))+e^(-3 x) ( dy(x))/( dx) = e^(-x)
Apply the reverse product rule f ( dg)/( dx)+g ( df)/( dx) = (d)/( dx)(f g) to the left-hand side:
( d)/( dx)(e^(-3 x) y(x)) = e^(-x)
Integrate both sides with respect to x:
integral ( d)/( dx)(e^(-3 x) y(x)) dx = integral e^(-x) dx
Evaluate the integrals:
e^(-3 x) y(x) = c_1-e^(-x), where c_1 is an arbitrary constant.
Divide both sides by mu(x) = e^(-3 x):
y(x) = e^(2 x) (c_1 e^x-1)
Solve for c_1 using the initial conditions:Substitute y(0) = 0 into y(x) = e^(2 x) (c_1 e^x-1):
c_1-1 = 0
Solve the equation:
c_1 = 1
Substitute c_1 = 1 into y(x) = e^(2 x) (c_1 e^x-1):
y(x) = e^(2 x) (e^x-1)

Solve
( du(v))/( dv) = (2 k)/v+v^3:
Integrate both sides with respect to v:
u(v) = integral ((2 k)/v+v^3) dv = c_1+2 k log(v)+v^4/4, where c_1 is an arbitrary constan
Solve the linear equation
x^2 ( dy(x))/( dx)+2 xy(x)-x+1 = 0

, such that y(1) = 0:
Subtract 1-x from both sides and divide by x^2:
(2 y(x))/x+( dy(x))/( dx) = -(1-x)/x^2
Let mu(x) = exp( integral 2/x dx) = x^2.
Multiply both sides by mu(x):
x^2 ( dy(x))/( dx)+(2 x) y(x) = x-1
Substitute 2 x = ( d)/( dx)(x^2):
y(x) ( d)/( dx)(x^2)+x^2 ( dy(x))/( dx) = x-1
Apply the reverse product rule f ( dg)/( dx)+g ( df)/( dx) = ( d)/(dx)(f g) to the left-hand side:
( d)/( dx)(x^2 y(x)) = x-1
Integrate both sides with respect to x:
integral ( d)/( dx)(x^2 y(x)) dx = integral (x-1) dx
Evaluate the integrals:
x^2 y(x) = c_1+x^2/2-x, where c_1 is an arbitrary constant.
Divide both sides by mu(x) = x^2:
y(x) = c_1/x^2-1/x+1/2
Solve for c_1 using the initial conditions:
Substitute y(1) = 0 into y(x) =c_1/x^2-1/x+1/2:
c_1-1/2 = 0
Solve the equation:
c_1 = 1/2
Substitute c_1 = 1/2 into y(x) = c_1/x^2-1/x+1/2:
y(x) = (x-1)^2/(2 x^2)
Solve
(y+x) ( dy)/( dx) = y-x:

y(x) = x v(x), entonces ( dy(x))/( dx) = v(x)+x ( dv(x))/( dx):
(x v(x)+x) (v(x)+x ( dv(x))/( dx)) = x v(x)-x
Simplify:
x (v(x)+1) (v(x)+x ( dv(x))/( dx)) = x (v(x)-1)
ahora ( dv(x))/( dx):
( dv(x))/( dx) = (-v(x)^2-1)/(x(v(x)+1))
Divide (-v(x)^2-1)/(v(x)+1):
((v(x)+1) ( dv(x))/( dx))/(-v(x)^2-1) = 1/x
Integrate both sides with respect to x:
integral ((v(x)+1) ( dv(x))/( dx))/(-v(x)^2-1) dx = integral 1/x dx
Evaluate the integrals:
-1/2 log(v(x)^2+1)-tan^(-1)(v(x)) = c_1+log(x), where c_1 is an arbitrary constant.
Substitute back for y(x) = x v(x):
-1/2 log(y(x)^2/x^2+1)-tan^(-1)((y(x))/x) =c_1+log(x)
Solve the linear equation
x^3 ( dy)/( dx)+2 x^2 y+-x^2+x = o,

such that y(1) = 0:
Subtract -x^2+x from both sides and divide by x^3:
(2 y(x))/x+( dy)/( dx) = (o+x^2+-x)/x^3
u(x) = exp( integral 2/x dx) = x^2.
entonces
x^2 ( dy(x))/( dx)+(2 x) y(x) = (o+x^2+-x)/x
Substitute 2 x = ( d)/( dx)(x^2):
y(x) ( d)/( dx)(x^2)+x^2 ( dy(x))/( dx) = (o+x^2+-x)/x
entonces f (dg)/( dx)+g ( df)/( dx) = ( d)/( dx)(f g entonces
( d)/( dx)(x^2 y(x)) = (o+x^2+-x)/x
Integrate respect to x:
integral ( d)/( dx)(x^2 y(x)) dx = integral (o+x^2+-x)/x dx
entonces:
x^2 y(x) = c_1+o log(x)+x^2/2-x, where c_1 is ahora u(x) = x^2:
y(x) = (c_1+o log(x)+x^2/2-x)/x^2
ahora condiciones iniciales:
Substitute y(1) = 0 into y(x) = (c_1+o log(x)+x^2/2-x)/x^2:
c_1-1/2= 0
Solve the equation:
c_1 = 1/2
Substitute c_1 = 1/2 into y(x) = (c_1+o log(x)+x^2/2-x)/x^2:
y(x) = (2 o log(x)+(x-1)^2)/(2 x^2)

c. El tanque queda vacio cuando h = 0.
o=-92005t+64
t=636 seg
Solve
v(dv/du)=(cosu senu - uv^2)/(u^2-1); v(0)=2
:
Let v(u) = u w(u), which gives ( dv(u))/( du) = w(u)+u ( dw(u))/( du):
u w(u) (w(u)+u ( dw(u))/( du)) = (-u^3 w(u)^2+sin(u)...