How do you solve the differential #dy/dx=(x-4)/sqrt(x^2-8x+1)#?

Answer 1

# y = sqrt(x^2-8x+1) + C #

#d y/dx=(x-4)/sqrt(x^2-8x+1) #

Is a First Order separable DE which we can sole by integrating:

# :. y = int \ (x-4)/sqrt(x^2-8x+1) \ dx #
Let # u=x^2-8x+1 => (du)/dx=2x-8 = 2(x-4) #

Substituting into the RHS integral we get:

# y = int \ (1/2)/sqrt(u) \ du # # \ \ = 1/2 int \ u^(-1/2) \ du # # \ \ = 1/2 u^(1/2)/(1/2) + C# # \ \ = sqrt(u) + C # # \ \ = sqrt(x^2-8x+1) + C #

which is the General Solution

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Answer 2

To solve the differential equation ( \frac{{dy}}{{dx}} = \frac{{x - 4}}{{\sqrt{{x^2 - 8x + 1}}}} ), you can use separation of variables.

  1. Separate variables: ( \frac{{dy}}{{\sqrt{{y}}}} = \frac{{x - 4}}{{\sqrt{{x^2 - 8x + 1}}}} , dx )

  2. Integrate both sides: ( \int \frac{{dy}}{{\sqrt{{y}}}} = \int \frac{{x - 4}}{{\sqrt{{x^2 - 8x + 1}}}} , dx )

  3. Solve the integrals: ( 2\sqrt{{y}} = \int \frac{{x - 4}}{{\sqrt{{x^2 - 8x + 1}}}} , dx )

  4. Simplify and solve for ( y ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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