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What is the solution of the Homogeneous Differential Equation? : #dy/dx = (x^2+y^2-xy)/x^2# with #y(1)=0#
1 Answer
Explanation:
We have:
# dy/dx = (x^2+y^2-xy)/x^2 # with #y(1)=0#
Which is a First Order Nonlinear Ordinary Differential Equation. Let us attempt a substitution of the form:
# y = vx #
Differentiating wrt #x# and applying the product rule, we get:
# dy/dx = v + x(dv)/dx #
Substituting into the initial ODE we get:
# v + x(dv)/dx = (x^2+(vx)^2-x(vx))/x^2 #
# y = vx #
Differentiating wrt #x# and applying the product rule, we get:
# dy/dx = v + x(dv)/dx #
Substituting into the initial ODE we get:
# v + x(dv)/dx = (x^2+(vx)^2-x(vx))/x^2 #
Alfred 2 8 download free. Then assuming that #x ne 0# this simplifies to:
# v + x(dv)/dx = 1+v^2-v #
# :. x(dv)/dx = v^2-2v+1 #
And we have reduced the initial ODE to a First Order Separable ODE, so we can collect terms and separate the variables to get: Busycal 3 1 6 download free.
# int 1/(v^2-2v+1) dv = int 1/x dx # Script center 1 0 download free.
# int 1/(v-1)^2 dv = int 1/x dx #
Both integrals are standard, so we can integrate to get:
# -1/(v-1) = ln|x| + C #
Using the initial condition, # y(1)=0 => v(1)=0 #, we get:
# -1/(0-1) = ln|1| + C => 1#
Thus we have:
# -1/(v-1) = ln|x| +1 #
# :. 1-v = 1/(1+ln|x|) #
# :. v = 1 - 1/(1+ln|x|) #
# = (1+ln|x|-1)/(1+ln|x|) #
# = (ln|x|)/(1+ln|x|) #
Then, we restore the substitution, to get the General Solution:
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# y/x = (ln|x|)/(1+ln|x|) #
# :. y = (xln|x|)/(1+ln|x|) #
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