# Solve the differential equation #x y'-y=x/sqrt(1+x^2)# ?

This is a linear nonhomogeneous differential equation. The solution is obtained as the sum of the homogeneous solution

and the particular solution

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To solve the differential equation (xy' - y = \frac{x}{\sqrt{1 + x^2}}), we can first rewrite it in a more standard form for a linear first-order differential equation. Let's denote (y' = \frac{dy}{dx}). Then, the equation becomes:

[xy' - y = \frac{x}{\sqrt{1 + x^2}}.]

Multiplying through by (\frac{1}{x}) to make the coefficient of (y') equal to 1, we get:

[y' - \frac{y}{x} = \frac{1}{\sqrt{1 + x^2}}.]

This is now in the standard form (y' + P(x)y = Q(x)), where (P(x) = -\frac{1}{x}) and (Q(x) = \frac{1}{\sqrt{1 + x^2}}). The integrating factor for this equation is (e^{\int P(x) , dx}), which in this case is (e^{-\int \frac{1}{x} , dx} = e^{-\ln|x|} = \frac{1}{|x|}).

Multiplying both sides of the equation by the integrating factor gives:

[\frac{1}{|x|}y' - \frac{1}{|x|}\frac{y}{x} = \frac{1}{|x|}\frac{1}{\sqrt{1 + x^2}}.]

This simplifies to:

[\frac{d}{dx}\left(\frac{y}{|x|}\right) = \frac{1}{|x|}\frac{1}{\sqrt{1 + x^2}}.]

Integrating both sides with respect to (x) gives:

[\frac{y}{|x|} = \int \frac{1}{|x|}\frac{1}{\sqrt{1 + x^2}} , dx + C,]

where (C) is the constant of integration. The integral on the right can be evaluated by letting (u = \sqrt{1 + x^2}), which gives (du = \frac{x}{\sqrt{1 + x^2}} , dx), leading to (\int \frac{1}{|x|}\frac{1}{\sqrt{1 + x^2}} , dx = \int \frac{1}{u} , du = \ln|u| + K = \ln|\sqrt{1 + x^2}| + K = \ln(1 + x^2)^\frac{1}{2} + K = \frac{1}{2}\ln(1 + x^2) + K).

Therefore, the solution to the differential equation is:

[y(x) = |x|\left(\frac{1}{2}\ln(1 + x^2) + K\right).]

where (K) is the constant of integration.

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To solve the given differential equation (xy' - y = \frac{x}{\sqrt{1+x^2}}), we can use the method of integrating factors.

First, we rewrite the equation in the form (y' + P(x)y = Q(x)), where (P(x) = -\frac{1}{x}) and (Q(x) = \frac{x}{x^2+1}).

Next, we find the integrating factor (I(x)) by taking the exponential of the integral of (P(x)): [I(x) = e^{\int P(x) , dx} = e^{\int -\frac{1}{x} , dx} = e^{-\ln|x|} = \frac{1}{|x|}]

Now, we multiply both sides of the differential equation by the integrating factor (I(x)): [\frac{1}{|x|} \cdot (xy' - y) = \frac{1}{|x|} \cdot \frac{x}{\sqrt{1+x^2}}]

Simplify: [y' - \frac{1}{|x|}y = \frac{1}{\sqrt{1+x^2}}]

Now, we integrate both sides: [\int y' , dx - \int \frac{1}{|x|}y , dx = \int \frac{1}{\sqrt{1+x^2}} , dx]

The first term integrates to (y), and the second term integrates to (\ln|x|) (assuming (x) is positive, if (x) is negative, the solution will involve absolute values). The right side integrates to (\sinh^{-1}(x)).

So, the general solution to the differential equation is: [y = \ln|x| + C \sinh^{-1}(x)] Where (C) is the constant of integration.

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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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