\(\frac{d\mathrm x}{dt}+2y = 5e^t\)

\(\frac{dy}{dt}-2\mathrm x=5e^t\)

By taking \(\frac{d}{dt}\) = D, we get

Dx + 2y = 5et ........(1)

Dy – 2x = 5et .........(2)

Now, operate D on both sides of equation (1), we get

D²x + 2Dy = 5De^t = 5e^t  .........(3)   \((\because De^t = \frac{d}{dt}e^t = e^t)\)

Now, multiply equation (2) by 2, we get

2Dy – 4x = 10e^t ............(4)

By subtracting equation (4) from equation (3), we get

D²x + 4x = –5e^t.

⇒ (D² + 4)x = –5e^t. ........(5)

Its auxiliary equation is m² + 4 = 0

⇒ m = \(\pm\)2i

Therefore, complementary function is

x = C₁cos2t + C₂sin2t

Particular integral is x \(=\frac{1}{D^2+4}(-5e^t)\)

\(\Rightarrow \mathrm x=\frac{-5e^t}{1^2+4}\) \(=\frac{-5e^t}{1+4}\) \(=\frac{-5e^t}{5}\) \(=-e^t\)

Hence, complete solution of equation (5) is

x = C.F + P.I = C₁cos2t + C₂sin2t – \(e^t\)

\(\Rightarrow \frac{d\mathrm x}{dt}\) = –2C₁sin2t + 2C₂cos2t – \(e^t\)

\(\Rightarrow\) 2y = 5e^t – \(\frac{d\mathrm x}{dt}\)  (From equation (1))

= 5e^t –(–2C₁sin2t + 2C₂cos2t – e^t)

= 6e^t + 2C₁sin2t – 2C₂cos2t

⇒ y = 3e^t + C₁sin2t – C₂cos2t

Hence, the solution of given system of equation is

x = C₁cos2t + C₂sin2t – e^t .........(6)

And y = C₁sin2t – C₂cos2t + 3e^t .......(7)

Given that when t = 0, then x = –1 and y = 3

Then from equation (6) and (7), we get

–1 = C₁ – 1 ⇒ C₁ = –1 + 1 = 0

And 3 = –C₂ + 3 ⇒ C₂ = 3 – 3 = 0

Hence, the solution of given system of equations is

x = –e^t 

And y = 3e^t  (By putting C₁ = 0 & C₂ = 0 in equations (6) and (7))