Solve the differential equation #xyy' +xyy'= y^2 +1#?

Question

Ezekiel Alexander · Accepted Answer

# y = +-sqrt(Bx -1)#

We have: # xyy' +xyy'= y^2 +1 # # :. 2xyy' = y^2 +1 # # :. y/(y^2 +1 )y' = 1/(2x)# This is a First Order separable DE, so we can separate the variables to get; # int \ y/(y^2 +1 ) \ dy = int \ 1/(2x) \ dx# The RHS is trivial and for the LHS we can use a substitution: Let #u=y^2+1 => (du)/dy = 2y #, or # int 1/2 ... \ du = int ... y\ dy # Substituting we get # int \ (1/2)/(u) \ du = int \ 1/(2x) \ dx# We can now integrate to get: # 1/2ln|u| = 1/2 ln |x| + A# # :. ln|u| = ln |x| + 2A# # :. ln|u| = ln |x| + lnB# (say) # :. ln|u| = ln B|x| # # :. u = Bx # # :. y^2+1 = Bx # # :. y^2 = Bx -1# # :. y = +-sqrt(Bx -1)#

Charles Arocha · Answer

undefined To solve the differential equation $xy\frac{dy}{dx} + xy' = y^2 + 1$, we can first rewrite it in a more standard form. Let's divide both sides by $x$:

$$y\frac{dy}{dx} + y' = \frac{y^2 + 1}{x}$$

Now, let $v = y^2$. Then, $v' = 2y y'$. Substitute $v$ and $v'$ into the equation:

$$ \begin{split} y\frac{dy}{dx} + y' & = \frac{v + 1}{x} \
2y y' & = \frac{v + 1}{x} - y\frac{dy}{dx} \end{split} $$

Now, substitute $v = y^2$:

$$ 2y y' = \frac{y^2 + 1}{x} - y\frac{dy}{dx} $$

This is a separable differential equation. Rearrange terms to get all $y$ terms on one side and $x$ terms on the other:

$$ 2y y' + y\frac{dy}{dx} = \frac{y^2 + 1}{x} $$

Now, we can separate variables:

$$ \begin{split} 2y \, dy + y \, dy & = \left( y^2 + 1 \right) \frac{dx}{x} \
3y \, dy & = \left( y^2 + 1 \right) \frac{dx}{x} \end{split} $$

Integrate both sides:

$$ \begin{split} \int 3y \, dy & = \int \left( y^2 + 1 \right) \frac{dx}{x} \
\frac{3}{2} y^2 & = \ln|x| + C \end{split} $$

Where $C$ is the constant of integration. Solving for $y^2$:

$$ y^2 = \frac{2}{3} \ln|x| + C' $$

Taking the square root of both sides:

$$ y = \pm \sqrt{\frac{2}{3} \ln|x| + C'} $$

Where $C'$ is the constant of integration.