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Consider the initial value problem:
y" - 5y' + 6y = -5sin(2t); y(0) = -5, y'(0) = 2
Write the differential equation above as a first-order system for the unknowns u, v, where u=y and v is defined as in the lecture notes. You do not need to solve the system.

Answer :

kudzordzifrancis

Answer:

[tex]\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2[/tex]

Step-by-step explanation:

The given initial value problem is;

[tex]y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2[/tex]

Let

[tex]u(t)=y(t)----(2)\\v(t)=y'(t)----(3)[/tex]

Differentiating both sides of equation (1) with respect to [tex]t[/tex], we obtain:

[tex]u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)[/tex]

Differentiating both sides of equation (2) with respect to [tex]t[/tex] gives:

[tex]v'(t)=y''(t)----(6)[/tex]

From equation (1),

[tex]y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)[/tex]

Putting t=0 into equation (2) yields

[tex]u(0)=y(0)\\\implies u(0)=-5[/tex]

Also putting t=0 into equation (3)

[tex]u'(0)=y'(0)\\u'(0)=2[/tex]

The system of first order equations is:

[tex]\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2[/tex]

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