What is the solution of the Homogeneous Differential Equation? : #dy/dx = (x^2+y^2-xy)/x^2# with #y(1)=0#

1. Initial 1041 Return
2. Https 10.0.0 1
3. Why Is 0 0 1

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1 Answer

Explanation:

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3. Y = (xln x )/(1+ln x ) 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 Then assuming that x ne 0 this simplifies to: v.

We have:

# dy/dx = (x^2+y^2-xy)/x^2 # with #y(1)=0#

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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 #

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: https://soft-panel.mystrikingly.com/blog/can-you-get-roblox-on-android.

# int 1/(v^2-2v+1) dv = int 1/x dx #

# 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 # Scrutiny 8 2 2 esv.

# :. 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:

# y/x = (ln|x|)/(1+ln|x|) #

# :. y = (xln|x|)/(1+ln|x|) #

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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 #

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:

# int 1/(v^2-2v+1) dv = int 1/x dx #

# int 1/(v-1)^2 dv = int 1/x dx # Minecraft download windows 10 edition.

Both integrals are standard, so we can integrate to get:

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# -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# Meta movie 2 4 3 torrent.

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:

Why Is 0 0 1

# y/x = (ln|x|)/(1+ln|x|) #

# :. y = (xln|x|)/(1+ln|x|) #

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