proceed with solving this differential equation for �(�)*v*(*t*)

Let's proceed step-by-step to solve the differential equation:

\[
10 \frac{d}{dt}v(t) + 2v(t) = e^{at} + e^{bt}
\]

### Step 1: Homogeneous Solution
We first solve the homogeneous equation:

\[
10 \frac{d}{dt}v_h(t) + 2v_h(t) = 0
\]

This can be rewritten as:

\[
\frac{d}{dt}v_h(t) = -\frac{2}{10} v_h(t) = -0.2 v_h(t)
\]

This is a first-order linear differential equation, whose solution is of the form:

\[
v_h(t) = C e^{-0.2 t}
\]

where \( C \) is a constant that we will determine using initial conditions later.

### Step 2: Particular Solution
Now, we find a particular solution for the non-homogeneous equation:

\[
10 \frac{d}{dt}v_p(t) + 2v_p(t) = e^{at} + e^{bt}
\]

Since the right-hand side consists of exponentials \( e^{at} \) and \( e^{bt} \), we assume the particular solution to be of the form:

\[
v_p(t) = A e^{at} + B e^{bt}
\]

Substitute \( v_p(t) \) into the differential equation:

\[
10 \left(A a e^{at} + B b e^{bt} \right) + 2(A e^{at} + B e^{bt}) = e^{at} + e^{bt}
\]

This simplifies to:

\[
(10 a A + 2A) e^{at} + (10 b B + 2B) e^{bt} = e^{at} + e^{bt}
\]

Equating coefficients for \( e^{at} \) and \( e^{bt} \), we get two equations:

1. \( (10a + 2)A = 1 \)
2. \( (10b + 2)B = 1 \)

### Step 3: Solve for A and B
From these equations:

1. \( A = \frac{1}{10a + 2} \)
2. \( B = \frac{1}{10b + 2} \)

### Step 4: General Solution
The general solution \( v(t) \) is the sum of the homogeneous and particular solutions:

\[
v(t) = C e^{-0.2t} + \frac{1}{10a + 2} e^{at} + \frac{1}{10b + 2} e^{bt}
\]

### Step 5: Applying Initial Conditions (if provided)
To fully determine \( C \), we would need an initial condition, such as the value of \( v(t) \) at \( t = 0 \). If no initial condition is given, we leave the solution in terms of \( C \).

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proceed with solving this differential equation for v(t)

Step-by-step solution to solving a first-order linear non-homogeneous differential equation with exponential functions. The problem involves finding both the homogeneous and particular solutions using key concepts from differential equations. Suitable for undergraduate students.

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