What is the (d^2-4)y(d,t)=5e^{3t},y(0)=1 ?

The (d^2-4)y(d,t)=5e^{3t},y(0)=1 is \circ d^2-4\circ

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(d^2-4)y(d,t)=5e^{3t},y(0)=1

(d^{2} - 4) y (d, t) = 5 e^{3 t}, y (0) = 1

\circ d^{2} - 4 \circ

Solution steps

Find

(d^{2} - 4) y (d, t) \circ y

given

(d^{2} - 4) y (d, t) (t) = 5 e^{3 t}, y (0) = 1

(d^{2} - 4) y (d, t) \circ y = (d^{2} - 4) y (d, t) \circ y (d)

y (d, t) : 1

y (d) : 1

= (d^{2} - 4) \cdot 1 \circ 1

Expand

(d^{2} - 4) \cdot 1 \circ 1 : \circ d^{2} - 4 \circ

= \circ d^{2} - 4 \circ

y^{^{''}} - 3 y^{^{'}} = 2 e^{x} sin (x)

y^{^{''}} - y^{^{'}} + y = sin (t)

y^{^{''}} + 2 y^{^{'}} + 3 y = 12 e^{x} + 3 x

y^{^{''}} - y = (1 - e^{2 x})^{- \frac{3}{2}}

y^{^{''}} + 4 y^{^{'}} - 2 y = x^{3}

What is the (d^2-4)y(d,t)=5e^{3t},y(0)=1 ?
The (d^2-4)y(d,t)=5e^{3t},y(0)=1 is \circ d^2-4\circ