Given a second order linear homogeneous differential equation a2(x)y+a1(x)y'+a0(x)y=0 a 2 (x) y + a...

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Given a second order linear homogeneous differential equation a2(x)y"+a1(x)y'+a0(x)y=0 a 2 (x) y " + a 1(x) y' + a0 (x) y = 0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions y1,y2 y 1, y 2. But there are times when only one function, call it y1 y 1, is available and we would like to find a second linearly independent solution. We can find y2 y 2 using the method of reduction of order. First, under the necessary assumption the a2(x)#0 a 2 ( x ) #0 we rewrite the equation as y"+p(x)y'+q(x)y=0 p(x)=a1(x)a2(x), q(x)=a0(x)a2(x), y" +p(x) y'+q (x)y=0p( x) = a1 (x) a 2 (x), q (x) = a0 (x) a 2 (x), Then the method of reduction of order gives a second linearly independent solution as y2(x)=Cylu=Cy1(x) [e-Sp(x)dxy21(x)dx y 2 (x) = Cylu=Cy1(x) [e- Sp(x) dxy12(x) dx where C C is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose C C so that all the constants in front reduce to 11. For example, if we obtain y2=C3e2x y 2 = C 3 e 2 x then we can choose C=1/3 C = 1/3 so that y2-e2x y 2 = e 2 x. Given the problem 25y"-20y'+4y=0 25 y" - 20 y' + 4y = 0 and a solution y1=e(2x/5) y 1 = e( 2x/5). Applying the reduction of order method to this problem we obtain the following y21(x)= y 12 (x) = p(x)= p (x) = and e-Sp(x)dx= e-S p(x) dx = So we have fe-Sp(x)dxy21(x)dx=[[e-Sp(x) dx y 12 (x) dx = [dx= d x = Finally, after making a selection of a value for C C as described above (you have to choose some nonzero numerical value) we arrive at y2(x)=Cylu= y 2 ( x ) = Cy1u = So the general solution to 25y"-20y'+4y=0 25 y" - 20 y' + 4y = 0 can be written as y=c1y1+c2y2=c1y = c1y 1+ c2y2=c1+c2 + c 2 Given a second order linear homogeneous differential equation a2(x)y"+a1(x)y'+a0(x)y=0 a 2 (x) y " + a 1(x) y' + a0 (x) y = 0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions y1,y2 y 1, y 2. But there are times when only one function, call it y1 y 1, is available and we would like to find a second linearly independent solution. We can find y2 y 2 using the method of reduction of order. First, under the necessary assumption the a2(x)#0 a 2 ( x ) #0 we rewrite the equation as y"+p(x)y'+q(x)y=0 p(x)=a1(x)a2(x), q(x)=a0(x)a2(x), y" +p(x) y'+q (x)y=0p( x) = a1 (x) a 2 (x), q (x) = a0 (x) a 2 (x), Then the method of reduction of order gives a second linearly independent solution as y2(x)=Cylu=Cy1(x) [e-Sp(x)dxy21(x)dx y 2 (x) = Cylu=Cy1(x) [e- Sp(x) dxy12(x) dx where C C is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose C C so that all the constants in front reduce to 11. For example, if we obtain y2=C3e2x y 2 = C 3 e 2 x then we can choose C=1/3 C = 1/3 so that y2-e2x y 2 = e 2 x. Given the problem 25y"-20y'+4y=0 25 y" - 20 y' + 4y = 0 and a solution y1=e(2x/5) y 1 = e( 2x/5). Applying the reduction of order method to this problem we obtain the following y21(x)= y 12 (x) = p(x)= p (x) = and e-Sp(x)dx= e-S p(x) dx = So we have fe-Sp(x)dxy21(x)dx=[[e-Sp(x) dx y 12 (x) dx = [dx= d x = Finally, after making a selection of a value for C C as described above (you have to choose some nonzero numerical value) we arrive at y2(x)=Cylu= y 2 ( x ) = Cy1u = So the general solution to 25y"-20y'+4y=0 25 y" - 20 y' + 4y = 0 can be written as y=c1y1+c2y2=c1y = c1y 1+ c2y2=c1+c2 + c 2

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