y^{''}+y=x,y(1)=0,y^'(1)=1

Lösung

y^{^{''}} + y = x, y (1) = 0, y^{^{'}} (1) = 1

y = \frac{( sin ^{2} ( 1 ) sec ( 1 ) - ( sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 ) ) ) cos ( x )}{cos ( 1 ) ( sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 ) )} - \frac{sec ( 1 ) sin ( 1 ) sin ( x )}{sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 )} + x

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{( sin ^{2} ( 1 ) sec ( 1 ) - ( sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 ) ) ) cos ( x )}{cos ( 1 ) ( sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 ) )} - \frac{sec ( 1 ) sin ( 1 ) sin ( x )}{sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 )} + x

y = \frac{( sin ^{2} ( 1 ) sec ( 1 ) - ( sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 ) ) ) cos ( x )}{cos ( 1 ) ( sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 ) )} - \frac{sec ( 1 ) sin ( 1 ) sin ( x )}{sin ^{2} ( 1 ) sec ( 1 ) + cos ( 1 )} + x

\int 3 x^{2} d x

\frac{\partial}{\partial y} (x cos (z))

t an g e n t 7 x^{2} - 9 x, at x = - 17

t an g e n t f (x) = 2 x^{2} - 5 x + 3, at x = 1

\int \frac{( 3 + 4 x )}{3 x - x ^{2} - 2} d x