Question

Find the value of b for which the equation
(y e^2xy + x) dx + bx e^2xy dy = 0
is exact, and hence solve it for that value of b .

Answer 1

Answer on Question #50670 – Math - Differential Calculus | Equations

Find the value of $b$ for which the equation

$$(y e ^ {\wedge} 2 x y + x) d x + b x e ^ {\wedge} 2 x y d y = 0$$

is exact, and hence solve it for that value of $b$.

Solution

$$(y e ^ {2 x y} + x) + b x e ^ {2 x y} \frac {d y}{d x} = 0$$

If the differential equation is exact, then

$$\frac {\partial}{\partial x} (b x e ^ {2 x y}) = \frac {\partial}{\partial y} (y e ^ {2 x y} + x) \rightarrow b e ^ {2 x y} + 2 b x y e ^ {2 x y} = e ^ {2 x y} + 2 x y e ^ {2 x y} \rightarrow$$ $$\rightarrow \quad b (1 + 2 x y) = 1 + 2 x y \rightarrow b = 1.$$

So, the exact equation is the following:

$$(y e ^ {2 x y} + x) + x e ^ {2 x y} \frac {d y}{d x} = 0$$

Solution of this exact equation is $\varphi (x,y) = C$,

where $\frac{\partial\varphi}{\partial x} = ye^{2xy} + x$, $\frac{\partial\varphi}{\partial y} = xe^{2xy}$, $C$ is an arbitrary real constant.

Thus, $\varphi (x,y) = \int xe^{2xy}dy + h(x) = \frac{e^{2xy}}{2} +h(x).$

From this formula and the previous one obtain

$$\frac {\partial \varphi}{\partial x} = y e ^ {2 x y} + h ^ {\prime} (x) = y e ^ {2 x y} + x \rightarrow h (x) = \frac {x ^ {2}}{2},$$ $$\varphi (x, y) = \frac {e ^ {2 x y}}{2} + \frac {x ^ {2}}{2}.$$

The solution of the equation is $e^{2xy} + x^2 = C_1$, hence $y = \frac{\ln(C_1 - x^2)}{2x}$, $x \neq 0$.

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