Find the particular solution of the differential equation dy/dx + 7y = 3 satisfy

Question

Find the particular solution of the differential equation
dy/dx + 7y = 3
satisfying the initial condition y(0) = 0 .
y=?

Explanation / Answer

First solve homogeneous problem: y' + 3y = 0 --> find general solution y(x) = exp(kx) y'(x) = k * exp(kx) --> k + 3 = 0 --> k = -3 y = C * exp(-3x) Now go back and find a particular solution to the inhomogeneous part. Try a solution that looks like the inhomogeneous part: ... = 7 a constant, guess a constant: y = D y' = 0 0 + 3 * D = 7 --> D = 7/3 Now add the inhomogeneous solution to the general solution to the homogeneous part: y = C * exp(-3x) + 7/3 Now put in your initial condition: x = 0, y = 1 1 = C * exp(-3*0) + 7/3 --> 1 = C + 7/3 --> C = 1 - 7/3 = -4/3 --> y(x) = (7 - 4 * exp(-3x)) / 3 Edit: If this is for Calculus and NOT Diff. EQ., then you can actually separate then integrate: dy/dx = 7 - 3y --> dy / (3y - 7) = -dx --> integrate both sides 1/3 * ln|3y - 7| = -x + C --> multiply by sides by 3 ln|3y - 7| = -3x + C --> take exponential of both sides 3y - 7 = C * exp(-3x) --> solve for y y = C * exp(-3x) + 7/3 --> exactly what we had before you can proceed from the steps there. (notice that C changed SEVERAL times, for instance 3C --> C because 3 times a constant is STILL a constant, so why write 3C. exp(-3x + C) = exp(C) * exp(-3x) --> exp(C) = a constant so just write C, not exp(C)...then finally C/3 --> C

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