Question

Solve: $(x^{2} + y^{2})dx + xy(d(x+y)) = 0$

Answer 1

$\ln(x^2(x^2+xy+2y^2)) + \frac{2}{\sqrt{7}}\arctan\left(\frac{2x+y}{y\sqrt{7}}\right) = C$

Answer 2

The given differential equation is $(x^{2} + y^{2})dx + xy(d(x+y)) = 0$.

First, expand $d(x+y)$ as $dx+dy$: $(x^{2} + y^{2})dx + xy(dx + dy) = 0$ $(x^{2} + y^{2})dx + xydx + xydy = 0$ $(x^{2} + xy + y^{2})dx + xydy = 0$

This is a homogeneous differential equation because both $M(x,y) = x^2+xy+y^2$ and $N(x,y) = xy$ are homogeneous functions of degree 2.

To solve a homogeneous differential equation, we can use the substitution $x = vy$. Then, $dx = vdy + ydv$.

Substitute $x = vy$ and $dx = vdy + ydv$ into the equation: $((vy)^2 + (vy)y + y^2)(vdy + ydv) + (vy)y dy = 0$ $(v^2y^2 + vy^2 + y^2)(vdy + ydv) + vy^2 dy = 0$ Factor out $y^2$: $y^2(v^2 + v + 1)(vdy + ydv) + vy^2 dy = 0$ Divide by $y^2$ (assuming $y \neq 0$): $(v^2 + v + 1)(vdy + ydv) + vdy = 0$ Distribute the terms: $v(v^2 + v + 1)dy + y(v^2 + v + 1)dv + vdy = 0$ $(v^3 + v^2 + v)dy + y(v^2 + v + 1)dv + vdy = 0$ Combine terms with $dy$: $(v^3 + v^2 + v + v)dy + y(v^2 + v + 1)dv = 0$ $(v^3 + v^2 + 2v)dy + y(v^2 + v + 1)dv = 0$ $v(v^2 + v + 2)dy + y(v^2 + v + 1)dv = 0$

Separate the variables: $\frac{dy}{y} = - \frac{v^2 + v + 1}{v(v^2 + v + 2)}dv$

Integrate both sides: $\int \frac{dy}{y} = - \int \frac{v^2 + v + 1}{v(v^2 + v + 2)}dv$

For the integral on the right-hand side, use partial fraction decomposition: Let $\frac{v^2 + v + 1}{v(v^2 + v + 2)} = \frac{A}{v} + \frac{Bv + C}{v^2 + v + 2}$ Multiply by $v(v^2 + v + 2)$: $v^2 + v + 1 = A(v^2 + v + 2) + (Bv + C)v$ $v^2 + v + 1 = Av^2 + Av + 2A + Bv^2 + Cv$ $v^2 + v + 1 = (A+B)v^2 + (A+C)v + 2A$

Comparing coefficients:

1. $2A = 1 \implies A = 1/2$
2. $A+C = 1 \implies 1/2 + C = 1 \implies C = 1/2$
3. $A+B = 1 \implies 1/2 + B = 1 \implies B = 1/2$

So, the integrand becomes: $\frac{1/2}{v} + \frac{1/2 v + 1/2}{v^2 + v + 2} = \frac{1}{2v} + \frac{v+1}{2(v^2 + v + 2)}$

Now, integrate: $\ln|y| = - \int \left( \frac{1}{2v} + \frac{v+1}{2(v^2 + v + 2)} \right) dv$ $\ln|y| = - \frac{1}{2} \int \frac{1}{v} dv - \frac{1}{2} \int \frac{v+1}{v^2 + v + 2} dv$

For the second integral, notice that the derivative of the denominator $v^2+v+2$ is $2v+1$. We can rewrite the numerator $v+1$ as $\frac{1}{2}(2v+2) = \frac{1}{2}(2v+1+1)$. So, $\int \frac{v+1}{v^2 + v + 2} dv = \int \frac{\frac{1}{2}(2v+1) + \frac{1}{2}}{v^2 + v + 2} dv = \frac{1}{2} \int \frac{2v+1}{v^2 + v + 2} dv + \frac{1}{2} \int \frac{1}{v^2 + v + 2} dv$

The first part of this integral is $\frac{1}{2}\ln|v^2+v+2|$. For the second part, complete the square in the denominator: $v^2 + v + 2 = v^2 + v + (\frac{1}{2})^2 - (\frac{1}{2})^2 + 2 = (v + \frac{1}{2})^2 + \frac{7}{4}$ So, $\frac{1}{2} \int \frac{1}{(v + \frac{1}{2})^2 + (\frac{\sqrt{7}}{2})^2} dv = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{7}}{2}} \arctan\left(\frac{v + \frac{1}{2}}{\frac{\sqrt{7}}{2}}\right) = \frac{1}{\sqrt{7}} \arctan\left(\frac{2v+1}{\sqrt{7}}\right)$

Putting it all together: $\ln|y| = -\frac{1}{2}\ln|v| - \frac{1}{2} \left[ \frac{1}{2}\ln|v^2+v+2| + \frac{1}{\sqrt{7}} \arctan\left(\frac{2v+1}{\sqrt{7}}\right) \right] + C_1$ $\ln|y| = -\frac{1}{2}\ln|v| - \frac{1}{4}\ln|v^2+v+2| - \frac{1}{2\sqrt{7}}\arctan\left(\frac{2v+1}{\sqrt{7}}\right) + C_1$

Multiply by 4: $4\ln|y| = -2\ln|v| - \ln|v^2+v+2| - \frac{2}{\sqrt{7}}\arctan\left(\frac{2v+1}{\sqrt{7}}\right) + 4C_1$ $\ln(y^4) = \ln(v^{-2}) + \ln((v^2+v+2)^{-1}) - \frac{2}{\sqrt{7}}\arctan\left(\frac{2v+1}{\sqrt{7}}\right) + C_2$ $\ln(y^4) = \ln\left(\frac{1}{v^2(v^2+v+2)}\right) - \frac{2}{\sqrt{7}}\arctan\left(\frac{2v+1}{\sqrt{7}}\right) + C_2$ $\ln(y^4 v^2 (v^2+v+2)) = C_2 - \frac{2}{\sqrt{7}}\arctan\left(\frac{2v+1}{\sqrt{7}}\right)$

Now substitute back $v = x/y$: $v^2 = x^2/y^2$ $v^2+v+2 = x^2/y^2 + x/y + 2 = (x^2+xy+2y^2)/y^2$ $y^4 \left(\frac{x^2}{y^2}\right) \left(\frac{x^2+xy+2y^2}{y^2}\right) = x^2(x^2+xy+2y^2)$

So the solution becomes: $\ln(x^2(x^2+xy+2y^2)) = C - \frac{2}{\sqrt{7}}\arctan\left(\frac{2(x/y)+1}{\sqrt{7}}\right)$ $\ln(x^2(x^2+xy+2y^2)) = C - \frac{2}{\sqrt{7}}\arctan\left(\frac{2x+y}{y\sqrt{7}}\right)$

The final answer is $\boxed{\ln(x^2(x^2+xy+2y^2)) + \frac{2}{\sqrt{7}}\arctan\left(\frac{2x+y}{y\sqrt{7}}\right) = C}$.