Question

Initial value problems dy/dx+2y=3. Y(0)=1

Answer 1

The initial value equation is $y = \frac 12(3 - e^{-2x})$

How to solve the initial value?

The equation is given as:

$\frac{dy}{dx} +2y = 3$

Where

Y(0) = 1

Subtract 2y from both sides in the equation

$\frac{dy}{dx} = -2y + 3$

Rewrite as:

$\frac{dy}{-2y + 3} = dx$

Integrate both sides of the equation

$-\frac 12\ln|-2y + 3| = x + C_o$

Multiply through by -2

$\ln|-2y + 3| = -2x + C_1$

Take the exponent of both sides

$-2y + 3 = Ce^{-2x$

Next, we solve for C under the initial condition Y(0) = 1.

This gives

$-2(1) + 3 = Ce^{-2 * 0$

Evaluate

-2 + 3 = C

Solve for C

C = 1

Substitute C = 1 in $-2y + 3 = Ce^{-2x$

$-2y + 3 = e^{-2x$

Next, we solve for y

$y = \frac 12(3 - e^{-2x})$

Hence, the initial value equation is $y = \frac 12(3 - e^{-2x})$

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