What is the solution for y^{''}+2y^'+1y=1,y(1)=1,y^'(1)=0 ?

The solution for y^{''}+2y^'+1y=1,y(1)=1,y^'(1)=0 is y=1

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y^{''}+2y^'+1y=1,y(1)=1,y^'(1)=0

y^{^{''}} + 2 y^{^{'}} + 1 y = 1, y (1) = 1, y^{^{'}} (1) = 0

y = 1

Solution steps

y^{''} (t) + 2 y^{'} (t) + 1 y = 1

Solve linear ODE:

y = 1

y = 1

n \to \infty lim (n^{2} x)

f (x) = e^{13}

y^{^{'}} - \frac{4}{x} y = \frac{y ^{4}}{x ^{4}}, y (1) = 1

\frac{\partial}{\partial x} (- x z)

\int ln (1 - 2 x) d x

What is the solution for y^{''}+2y^'+1y=1,y(1)=1,y^'(1)=0 ?
The solution for y^{''}+2y^'+1y=1,y(1)=1,y^'(1)=0 is y=1