Question

Solve the following ordinary differential equations
xdx+ydy+4y^3(x^2+y^2)dy

Answer 1

Answer on Question #39003 - <math> - <algebra>.

Solve the following ordinary differential equations; $xdx + ydy + 4y^3 (x^2 + y^2)dy$

Solution:

We can rewrite our equation:

$$x \, dx + \left(4(x^2 + y(x)^2) y(x)^3 + y(x)\right) d y(x) = 0$$

Make the replacement, let $R(x,y) = x$ and $S(x,y) = 4y^3(x^2 + y^2) + y$. This is not exact equation, because $\frac{\partial R(x,y)}{\partial y} = 0 \neq 8xy^3 = \frac{\partial S(x,y)}{\partial x}$.

Next step we have to find an integrating factor $\mu(y)$ such that

$$\mu(y) R(x,y) dx + \mu(y) S(x,y) d y(x) = 0 \text{ is exact}.$$

This means that $\frac{\partial}{\partial y} (\mu(y)R(x,y)) = \frac{\partial}{\partial x} (\mu(y)S(x,y))$:

$$\frac{\partial \mu(y)}{\partial y} x = 8 y^3 \mu(y) x$$

Isolate $\mu(y)$ to the left-hand side:

$$\frac{\frac{\partial \mu(y)}{\partial y}}{\mu(y)} = 8 y^3$$

Integrate both sides with respect to $y$:

$$\log \left(\mu(y)\right) = 2 y^4$$

Then we take exponentials of the both sides:

$$\mu(y) = e^{2 y^4}$$

Go back to our expression and multiply it:

$$\left(4(x^2 + y(x)^2) y(x)^3 + y(x)\right) \frac{dy(x)}{dx} + x = 0 \text{ by } \mu(y(x)):$$ $$e^{2 y(x)^4} x + \left(e^{2 y(x)^4} \left(4(x^2 + y(x)^2) y(x)^3 + y(x)\right)\right) \frac{dy(x)}{dx} = 0$$

Let $P(x,y) = x e^{2y^4}$ and $Q(x,y) = e^{2y^4} (4y^3(x^2 + y^2) + y)$ this is an exact equation, because $\frac{\partial P(x,y)}{\partial y} = 8x e^{2y^4} y^3 = \frac{\partial Q(x,y)}{\partial x}$.

Define $f(x,y)$ such that $\frac{\partial f(x,y)}{\partial x} = P(x,y)$ and $\frac{\partial f(x,y)}{\partial y} = Q(x,y)$:

Then the solution will be given by $f(x,y) = C_1$, where $C_1$ is an arbitrary constant. Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to $x$ in order to find $f(x,y)$:

$$f(x, y) = \int e^{2y^4} x \, dx = \frac{1}{2} e^{2y^4} x^2 + g(y)$$

Where $g(y)$ is an arbitrary function of $y$.

Differentiate $f(x, y)$ with respect to $y$ in order to find the $g(y)$:

$$\frac{\partial f(x, y)}{\partial y} = \frac{\partial}{\partial y} \left( \frac{1}{2} e^{2y^4} x^2 + g(y) \right) = 4 e^{2y^4} y^3 x^2 + \frac{\partial g(y)}{\partial y}$$

Substitute into $\frac{\partial f(x, y)}{\partial y} = Q(x, y)$:

$$4 e^{2y^4} y^3 x^2 + \frac{\partial g(y)}{\partial y} = e^{2y^4} \left( y + 4 y^3 (x^2 + y^2) \right)$$

Then we solve for $\frac{\partial g(y)}{\partial y}$:

$$\frac{\partial g(y)}{\partial y} = e^{2y^4} (4 y^5 + y)$$

Integrate $\frac{\partial g(y)}{\partial y}$ with respect to $y$:

$$g(y) = \int e^{2y^4} (4 y^5 + y) \, dy = \frac{1}{2} e^{2y^4} y^2$$

Substitute $g(y)$ into $f(x, y)$:

$$f(x, y) = \frac{1}{2} e^{2y^4} x^2 + \frac{1}{2} e^{2y^4} y^2$$

Eventually we get the solution $f(x, y) = C_1$ (as we note earlier).

The solution is $C_1 = \frac{1}{2} e^{2y^4} x^2 + \frac{1}{2} e^{2y^4} y^2$

$$C_1 = \frac{1}{2} x^2 e^{2y(x)^4} + \frac{1}{2} e^{2(x)^4} y(x)^2$$

Answer: $C_1 = \frac{1}{2} x^2 e^{2y(x)^4} + \frac{1}{2} e^{2(x)^4} y(x)^2$