derivative of (2y+3^4)=5x^3-3x

Lösung

\frac{d}{d x} ((2 y + 3)^{4}) = 5 x^{3} - 3 x

y = \frac{4 5 x ^{4} - 6 x ^{2} + c _{1}}{2 2} - \frac{3}{2}, y = - \frac{4 5 x ^{4} - 6 x ^{2} + c _{1}}{2 2} - \frac{3}{2}

Schritte zur Lösung

\frac{d}{d x} ((2 y + 3)^{4}) = 5 x^{3} - 3 x

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

(2 y + 3)^{4} = \int 5 x^{3} - 3 x d x

\int 5 x^{3} - 3 x d x = \frac{5 x ^{4}}{4} - \frac{3 x ^{2}}{2} + c_{1}

(2 y + 3)^{4} = \frac{5 x ^{4}}{4} - \frac{3 x ^{2}}{2} + c_{1}

Isoliere

y : y = \frac{4 5 x ^{4} - 6 x ^{2} + 4 c _{1}}{2 2} - \frac{3}{2}, y = - \frac{4 5 x ^{4} - 6 x ^{2} + 4 c _{1}}{2 2} - \frac{3}{2}

y = \frac{4 5 x ^{4} - 6 x ^{2} + 4 c _{1}}{2 2} - \frac{3}{2}, y = - \frac{4 5 x ^{4} - 6 x ^{2} + 4 c _{1}}{2 2} - \frac{3}{2}

Vereinfache

y = \frac{4 5 x ^{4} - 6 x ^{2} + c _{1}}{2 2} - \frac{3}{2}, y = - \frac{4 5 x ^{4} - 6 x ^{2} + c _{1}}{2 2} - \frac{3}{2}

\frac{d y}{d x} - 11 x^{3} y = 0

\frac{d x}{d t} = x \cdot sin (t)

x \frac{d y}{d x} = y^{2} - 1

(x^{2} + 37) y^{^{'}} = x y

(e^{x} + 1) \frac{d y}{d x} + y e^{x} - y = 0