5x(x+4y)y^'=5y(x-4y)

Lösung

5 x (x + 4 y) y^{^{'}} = 5 y (x - 4 y)

y = \frac{x}{4 W ( \frac{x ^{2}}{c _{1}} )}

Schritte zur Lösung

5 x (x + 4 y) y^{'} (x) = 5 y (x - 4 y)

Solve using substitution:

y = \frac{x}{4 W ( \frac{x ^{2}}{c _{1}} )}

y = \frac{x}{4 W ( \frac{x ^{2}}{c _{1}} )}

3 y^{^{'}} + 7 y = x^{3} - 2 x^{2} + 1

y^{^{'}} = x \cdot y - 3 \cdot x + 3 \cdot y + 3 \cdot y - 9

y^{^{'}} = y + e^{x}, y (0) = (x + c) e^{x}, y (0) = \frac{1}{2}

y^{^{'}} = (x + y) 2

y^{^{'}} = 2 y + (x^{2}) + 5