Solve the differential equation #2xlnx dy/dx + y = 0#?

Answer 1

# y = A/sqrt(lnx) #

We have:

# 2xlnx dy/dx + y = 0# ..... [A]

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

So rewrite the equations in standard form as:

# dy/dx + 1/(2xlnx)y = 0 ..... [B] #

Then the integrating factor is given by;

# I = e^(int P(x) dx) # # \ \ = exp(int \ 1/(2xlnx) \ dx) # # \ \ = exp( 1/2ln|lnx| ) # (see notes at end) # \ \ = exp( lnsqrt(|lnx|)) # # \ \ = sqrt(lnx)) #
And if we multiply the DE [B] by this Integrating Factor, #I#, we will have a perfect product differential form of #[A]#;
# sqrt(lnx)dy/dx + sqrt(lnx)1/(2xlnx)y = 0 # # :. d/dx (ysqrt(lnx)) = 0 #
# :. ysqrt(lnx)) = A #

Which we can rearrange to get:

# :. y = A/sqrt(lnx) #

Which, is the General Solution.

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Answer 2

The solution to the differential equation (2x \ln(x) \frac{dy}{dx} + y = 0) is given by:

[y = Cx^{-2}]

where (C) is an arbitrary constant.

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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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