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

Question

Scarlett Allison · Accepted Answer

#y = (C_2+arcsin(x))x#

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

#x y'_h-y_h=0# (1)

and the particular solution

#x y'_p-y_p=x/sqrt(1+x^2)# (2) so

#y = y_h + y_p# (3)

The homogeneous solution is #y_h=C_1x# The particular is obtained using the constant variation method due to Lagrange. So we make

#y_p=C(x)x# and substituting into (2) we obtain

#C'(x)=1/sqrt(1+x^2)#

integrating #C(x)# we get

#C(x)=arcsin(x)# and finally

#y = (C_2+arcsin(x))x#

Charles Arocha · Answer

undefined 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.

Emilia Aguilar · Answer

undefined 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.