One solution of the differential equation y\" - y\' = 0 is y = e^x. Use the meth
ID: 1944541 • Letter: O
Question
One solution of the differential equation y" - y' = 0 is y = e^x. Use the method of reduction of order to find a second linearly independent solution.Solve the differential equation
y" + 7y' + 12y = 0
Explanation / Answer
One solution of the differential equation y" - y' = 0 is y = e^x. let Y1 and Y2 be 2 solutions .here Y1 = e^x by reduction of order method take Y2 = a(x) * Y1 = a(x) * e^x substituting Y2 in diff equation y"-y' =0 as Y2 is also solution of that equation. => y'(x) = a'(x)*e^x+ e^x*a(x) , y"(x) = a"(x) *e^x + a'(x) * e^x + e^x*a(x)+ a'(x)*e^x = a"(x) *e^x +2 a'(x) * e^x +e^x*a(x) substituting in diff eqn y"-y' =0 we get a"(x) *e^x +2 a'(x) * e^x + e^x*a(x) -( a'(x)*e^x+ e^x*a(x) ) =0 =>a"(x) *e^x+ a'(x)*e^x = 0 =>e^x *(a"(x)+a'(x)) =0 =>a"(x)+a'(x) =0 => a(x) = C1 +C2 *x so solution Y2 = (C1 +C2 *x)*e^x y" + 7y' + 12y = 0 put y=e^mx then substituting in diff eqn we get m^2 + 7m + 12 =0 =>(m+3)*(m+4) =0 =>m=-3 ,-4 therefore solution is y =e^-3x or y=e^-4x or combined solution y = A *e^-3x + b*e^-4x
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