What is the solution for u^{''}+(a^2+b)\H(t)=0 ?

The solution for u^{''}+(a^2+b)\H(t)=0 is u(t)=integral of integral of-\H(t)(a^2+b)

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u^{''}+(a^2+b)\H(t)=0

u^{^{''}} + (a^{2} + b) H (t) = 0

u (t) = \int\int - H (t) (a^{2} + b) d t d t

Solution steps

u^{''} (t) + (a^{2} + b) H (t) = 0

Isolate

u^{''} (t) : u^{''} (t) = - H (t) (a^{2} + b)

Solve

u^{''} (t) = - H (t) (a^{2} + b) : u^{'} (t) = \int - H (t) (a^{2} + b) d t

Solve

u^{'} (t) = \int - H (t) (a^{2} + b) d t : u (t) = \int \int - H (t) (a^{2} + b) d t d t

u (t) = \int \int - H (t) (a^{2} + b) d t d t

y^{^{''}} + 13 y^{^{'}} + 36 y = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 7 y = 0

y^{^{''}} + 2 y^{^{'}} - 3 y = 0, y (0) = 2, y^{^{'}} (0) = 10

\frac{d ^{4} y}{d x ^{4}} + 6 \frac{d ^{2} y}{d x ^{2}} + 9 y = 0

y^{^{'''}} + (- y^{^{''}}) + 4 y^{^{'}} - 4 y = 0

What is the solution for u^{''}+(a^2+b)\H(t)=0 ?
The solution for u^{''}+(a^2+b)\H(t)=0 is u(t)=integral of integral of-\H(t)(a^2+b)