y^'=t+3

Lösung

y^{^{'}} = t + 3

y = \frac{t ^{2}}{2} + 3 t + c_{1}

Schritte zur Lösung

y^{'} (t) = t + 3

Wenn

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

dann

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

y = \int t + 3 d t

\int t + 3 d t = \frac{t ^{2}}{2} + 3 t + c_{1}

y = \frac{t ^{2}}{2} + 3 t + c_{1}

\frac{\partial}{\partial x} (\frac{( 3 x + 5 )}{6 y + 1})

\int_{0}^{2} π (4 x^{2} - x^{4}) d x

t an g e n t y = cos (x), at x = \frac{π}{4}

y^{^{''}} + y^{^{'}} - y = 0

\frac{\partial}{\partial x} (\frac{4 x - 4 y}{4 x + 4 y})