(dy/dx) + 8y =5
To find the solution to the given differential equation (dy/dx) + 8y = 5, we can use the method of solving first-order linear differential equations.
Step 1: Write the differential equation in standard form.
The given differential equation is already in standard form.
(dy/dx) + 8y = 5
Step 2: Identify the integrating factor.
The integrating factor can be found by multiplying the entire equation by the exponential of the integral of the coefficient of y, which in this case is 8. The integrating factor (IF) can be obtained as follows:
IF = e^(∫ 8 dx)
= e^(8x)
Step 3: Multiply the original equation by the integrating factor.
Now, multiply both sides of the differential equation by the integrating factor (IF), which is e^(8x):
e^(8x) * (dy/dx) + 8e^(8x) * y = 5e^(8x)
Step 4: Simplify the left-hand side.
The left-hand side can be simplified by applying the product rule of differentiation to the first term:
(e^(8x) * y)' + 8e^(8x) * y = 5e^(8x)
Step 5: Integrate both sides of the equation.
Integrate both sides of the equation with respect to x:
∫ [(e^(8x) * y)' + 8e^(8x) * y] dx = ∫ 5e^(8x) dx
Step 6: Simplify the integrals.
On the left-hand side of the equation, the integral of the derivative yields the original function:
e^(8x) * y + ∫ [8e^(8x) * y] dx = ∫ 5e^(8x) dx
e^(8x) * y + 8 ∫ [e^(8x) * y] dx = 5 ∫ e^(8x) dx
Step 7: Solve the integrals.
Evaluate the integrals on both sides of the equation:
e^(8x) * y + 8 ∫ [e^(8x) * y] dx = 5 ∫ e^(8x) dx
e^(8x) * y + 8 ∫ [e^(8x) * y] dx = (5/8) e^(8x) + C
Step 8: Solve for y.
To find y, solve the equation for y by isolating it on one side:
e^(8x) * y + 8 ∫ [e^(8x) * y] dx = (5/8) e^(8x) + C
Multiply through by e^(-8x):
y + 8 ∫ y dx = (5/8) + Ce^(-8x)
The integral term on the left-hand side can be solved as:
∫ y dx = yx + K
Substituting this back into the equation:
y + 8yx + 8K = (5/8) + Ce^(-8x)
Combine like terms:
y(1 + 8x) = (5/8) + Ce^(-8x) - 8K
Simplifying further:
y = [(5/8) + Ce^(-8x) - 8K] / (1 + 8x)
This is the solution to the given differential equation.