How to solve for P(x) if P'(x)=4P(x)+a, assuming a is a non-zero constant

Question

I have to solve this differential equation for my calculus class, and I've gotten stuck. I found that if P(x)=e^4x then P'(x)=4e^4x which I think is a step in the right direction, but I'm confused on how to deal with a here. How should I approach this problem and other problems like this?

Answer 1

Amongst other methods, you can solve this differential equation using separation of variables because it is algebraically possible to rearrange this differential equation with all the P terms on one side of the equation and all the x terms on the other.

To solve for P(x) using separation of variables, let's first call P'(x) = dP/dx , and then rewrite the original differential equation as:

dP/dx = 4P + a

Now, rearrange this such that all P terms are on one side of the equation, and all x terms on the other side:

dP/(4P+a)= dx

Now, integrate both sides:

∫dp/(4P+a) = ∫dx

which results in:

0.25ln(4P+a) + C₁ = x + C₂

Rearranging:

0.25ln(4P+a) = x + C₂ - C₁

For simplicity, we can replace C₂ - C₁ with another arbitrary constant C

0.25ln(4P+a) = x + C

Now, algebraically rearrange to solve for P:

P =(1/4)e^4Ce^4x - a/4

Because (1/4)e^4C is just some constant, we can replace it with arbitrary constant C for simplicity, yielding our final solution for P(x)

P(x) = Ce^4x - a/4

If you were given initial conditions, you could solve for the exact value of the constant C. Additionally, as a way to "check" that our P(x) is correct, you can always plug this P(x) solution back into the original differential equation to see if it holds up:

If we found that P(x) = Ce^4x - a/4, then our proposed P'(x) = 4Ce^4X

Now plug this into the original differential equation and simplify:

P'(x)=4P(x)+a

4Ce^4X = 4 (Ce^4x - a/4) + a

4Ce^4X = 4 Ce^4x - a + a

4Ce^4X = 4 Ce^4x

Hence, our solution P(x) = Ce^4x - a/4 is correct

Answer 2

P'(x) - 4P(x) = a; Integrated factor m = e^∫-4dx = e^-4x;

P'(x)e^-4x - 4P(x)e^-4x = ae^-4x; d/dx(P(x)e^-4x) = ae^-4x; P(x)e^-4x = ∫ae^-4xdx; P(x)e^-4x = - a/4e^-4x + C;

P(x) = - a/4 + Ce^4x

Answer 3

P' = 4P + a

Separable equation

P'/(4P+a) = 1

∫dP/(4P+a) = ∫dx

ln(|4P+a|)/4 = x + C₁

ln(|4P+a|) = 4x + C₂

4P+a = e^4x+C₂=C₃e^4x

P=C₄e^4x-a/4

C₄ and a will be determined by any initial conditions such as P(0) = ? and P'(0) = ??