First Order ODE/exp x (1 + x) dx = (x exp x - y exp y) dy
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Theorem
The first order ordinary differential equation:
- $(1): \quad e^x \paren {1 + x} \rd x = \paren {x e^x - y e^y} \rd y$
has the solution:
- $2 x e^x e^{-y} + y^2 + C$
Proof
Dividing $(1)$ through by $e^y$:
- $(2): \quad \paren {1 + x} e^x e^{-y} \rd x = \paren {x e^x e^{-y} - y} \rd y$
Let $z = x e^x e^{-y}$.
Then:
- $\dfrac {\d z} {\d x} = \paren {1 + x} e^x e^{-y} - x e^x e^{-y} \dfrac {\d y} {\d x}$
or:
- $\d z = \paren {1 + x} e^x e^{-y} \rd x - x e^x e^{-y} \rd y$
Substituting $z$ and $\d z$ in $(2)$:
- $\d z = \paren {z - y} \rd y - z \rd y = - y \rd y$
which directly leads to:
- $\ds \int \rd z = -\int y \rd y$
from which:
- $z = -\dfrac {y^2} 2 + k$
Substituting for $z$ and letting $C = 2 k$:
- $2 x e^x e^{-y} + y^2 + C$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $25$