u^'+n^2\H(t)=-((-1)^n+1)/(pi(n^2-1))

Lösung

u^{^{'}} + n^{2} H (t) = - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )}

u (t) = \int - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )} - n^{2} H (t) d t

Schritte zur Lösung

u^{'} (t) + n^{2} H (t) = - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )}

Isoliere

u^{'} (t) : u^{'} (t) = - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )} - n^{2} H (t)

Wenn

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

dann

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

u (t) = \int - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )} - n^{2} H (t) d t

u (t) = \int - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )} - n^{2} H (t) d t

y^{^{'}} - sin (x) \cdot y = 3 sin (x)

y^{2} y^{^{'}} = (1 + y^{3}) cos (x)

\frac{d y}{d t} = y (y - 1)

(x y + y^{2}) d x - (x^{2} + x y) d y = 0

y^{^{'}} = \frac{x ^{4}}{y ^{2}}