Question

please check if differential equations are solved correctly

dy/dx – 4y = e˟ given that y(0) = 0
using dy/dx + Py = Q
P = -4 & Q = e˟
Integrating factor = e˄∫-4dx
e¯⁴˟
solution:
(e¯⁴˟)y = ∫ (e¯⁴˟ )( e˟) dx
(e¯⁴˟)y = ∫ (e¯³˟ ) dx
(e¯⁴˟)y = -¼ e¯⁴˟ + C
y = -¼eº +C(e⁴˟)
y = -¼ +C(e⁴˟)

y(0) = -¼ +C(e⁴˟)
0 + ¼ = C(eº)
C = ¼
y = -¼ + ¼ e⁴˟

Answer 1

I would first solve the homogeneous equation
y' -4y = 0

using the characteristic equation m-4=0

then use the "method of undetermined coefficients" to find the particular solution
in this approach you assume y = A e^x
plug that into the diff eq and solve for A

Answer 2

if we use the integration factor, you are ok up through
solution:
(e¯⁴˟)y = ∫ (e¯⁴˟ )( e˟) dx
(e¯⁴˟)y = ∫ (e¯³˟ ) dx

the integral of (e¯³˟ ) dx is -1/3 e¯³˟
NOT -¼ e¯⁴˟
apparently you are thinking of the power rule of integration.
but the integral of e^x dx is e^x

in this case you want to integrate \( -\frac{1}{3}e^{-3x} (-3) dx \)

Answer 3

Thanks, fixed it as follows:
Solution:
(e¯⁴˟)y = ∫ (e¯⁴˟ )( e˟) dx
(e¯⁴˟)y = ∫ (e¯³˟ ) dx
(e¯⁴˟)y = -⅓ e¯³˟ + C
y = -⅓ (e⁴˟)(e¯³˟) + e⁴˟ C
y = -⅓ (e˟) +C(e⁴˟)

y(0) = -⅓ (e˟) +C(e⁴˟)
0 + ⅓ = C(eº)
C = ⅓
y = -⅓ + ⅓(e⁴˟)

Answer 4

yes, you can always check it by putting it into the original differential equation

Answer 5

although your last line left out e^x from the first term

Answer 6

thanks again.