The initial value equation is [tex]y = \frac 12(3 - e^{-2x})[/tex]
The equation is given as:
[tex]\frac{dy}{dx} +2y = 3[/tex]
Where
Y(0) = 1
Subtract 2y from both sides in the equation
[tex]\frac{dy}{dx} = -2y + 3[/tex]
Rewrite as:
[tex]\frac{dy}{-2y + 3} = dx[/tex]
Integrate both sides of the equation
[tex]-\frac 12\ln|-2y + 3| = x + C_o[/tex]
Multiply through by -2
[tex]\ln|-2y + 3| = -2x + C_1[/tex]
Take the exponent of both sides
[tex]-2y + 3 = Ce^{-2x[/tex]
Next, we solve for C under the initial condition Y(0) = 1.
This gives
[tex]-2(1) + 3 = Ce^{-2 * 0[/tex]
Evaluate
-2 + 3 = C
Solve for C
C = 1
Substitute C = 1 in [tex]-2y + 3 = Ce^{-2x[/tex]
[tex]-2y + 3 = e^{-2x[/tex]
Next, we solve for y
[tex]y = \frac 12(3 - e^{-2x})[/tex]
Hence, the initial value equation is [tex]y = \frac 12(3 - e^{-2x})[/tex]
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