(dy)/(dx)=4(x^2+1),x(pi/4)=1

Lösung

\frac{d y}{d x} = 4 (x^{2} + 1), x (\frac{π}{4}) = 1

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

Schritte zur Lösung

\frac{d y}{d x} = 4 (x^{2} + 1)

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

y^{^{'}} (x) = 4 (x^{2} + 1)

Wenn

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

dann

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

y = \int 4 (x^{2} + 1) d x

\int 4 (x^{2} + 1) d x = 4 (\frac{x ^{3}}{3} + x) + c_{1}

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

\frac{d}{d x} (1 - 4 cos (x))

\int_{- 2}^{2} \frac{1}{x ^{2} - 9} d x

\frac{d v}{d t} = 9.8 - v

\int \frac{x ( x - 8 )}{( x - 4 ) ^{3}} d x

x \to 1 lim (\frac{2 x - 7}{4 x + 5})