What is the solution of the Homogeneous Differential Equation? : #dy/dx = (x^2+y^2-xy)/x^2# with #y(1)=0#
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- Sep 10, 2017 First, split off your separate expressions into sub-functions. Let y=t.u.v where t=x^2+1, u=(x+2)^2, and v=(x-3)^3 Then dt/dx = 2x. (du)/dx = 2(x+2). By the chain rule, (dv)/dx = 3(x-3)^2. The product rule for three terms states: If y=t.u.v, and y is a function of x. Then dy/dx = dt/dx.u.v + (du)/dx.t.v + (dv)/dx.t.u. So, dy/dx = 2x.(x+2)^2.(x-3)^3+2(x^2+1).(x-3)^3.(x+2)+3(x^2 + 1).(x+2)^2.(x.
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1 Answer
Explanation:
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Which is a First Order Nonlinear Ordinary Differential Equation. Let us attempt a substitution of the form:
Differentiating wrt #x# and applying the product rule, we get:
Substituting into the initial ODE we get:
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Then assuming that #x ne 0# this simplifies to:
And we have reduced the initial ODE to a First Order Separable ODE, so we can collect terms and separate the variables to get:
Both integrals are standard, so we can integrate to get:
Using the initial condition, # y(1)=0 => v(1)=0 # , we get:
Thus we have:
Then, we restore the substitution, to get the General Solution:
Microsoft office download. # y/x = (ln|x|)/(1+ln|x|) #
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