Obtain the general solutions given the differential operators (D + 3)4 y = 0
4(D+3)y=0⇒(D+3)y=0⇒(ddx+3)y=0⇒dydx+3y=0⇒dydx=−3y⇒dyy=−3dx4(D + 3) y = 0 \\ \Rightarrow (D + 3) y = 0 \\ \Rightarrow (\dfrac d{dx} + 3) y = 0 \\ \Rightarrow \dfrac {dy}{dx}+3y=0 \\ \Rightarrow \dfrac {dy}{dx}=-3y \\ \Rightarrow \dfrac {dy}{y}=-3dx4(D+3)y=0⇒(D+3)y=0⇒(dxd+3)y=0⇒dxdy+3y=0⇒dxdy=−3y⇒ydy=−3dx
On integrating both sides,
logy=−3x+C⇒y=e−3x+C⇒y=C1e−3x, where C1=eC\log y=-3x+C \\ \Rightarrow y=e^{-3x+C} \\ \Rightarrow y=C_1e^{-3x}, \ where \ C_1=e^Clogy=−3x+C⇒y=e−3x+C⇒y=C1e−3x, where C1=eC
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment