Question

(2yx^2 +4)dy/dx+(2y^2x -3)=0

Answer 1

Answer on Question #74062 – Math – Differential Equations

$$(2yx^2 + 4) \frac{dy}{dx} + (2y^2x - 3) = 0$$

Solution

$$(2yx^2 + 4) \frac{dy}{dx} + (2y^2x - 3) = 0$$ $$(2yx^2 + 4)dy + (2y^2x - 3)dx = 0$$

Let $P(x,y) = 2yx^2 + 4$, $Q(x,y) = 2y^2x - 3$

This is an exact equation because

$$\frac{\partial P(x,y)}{\partial x} = \frac{\partial Q(x,y)}{\partial y} = 4xy$$ $$\frac{\partial F(x,y)}{\partial x} = 2y^2x - 3$$ $$\frac{\partial F(x,y)}{\partial y} = 2yx^2 + 4$$ $$\int \frac{\partial F(x,y)}{\partial x} dx = \int (2y^2x - 3)dx + \varphi(y)$$ $$F(x,y) = x^2y^2 - 3x + \varphi(y)$$ $$\frac{\partial (x^2y^2 - 3x + \varphi(y))}{\partial y} = 2yx^2 + 4$$ $$2yx^2 + \varphi'(y) = 2yx^2 + 4$$ $$\varphi'(y) = 4 \rightarrow \varphi(y) = \int 4dy = 4y$$

The solution is $F(x,y) = C$

$$x^2y^2 - 3x + 4y = C$$

Answer: $x^2y^2 - 3x + 4y = C$

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