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please explain in detailed steps the general solution of the following differential equation. given equation: y"-3y'/x=x solution: C1x^4/4-x^3/3+C2 required : detailed steps in the solution of the above
Dear student, Let y=e^mx dy/dx=me^mx 2y/dx2=m2e^mx Now substitute this and solve the equation......
Dear student,
Let y=e^mx
dy/dx=me^mx
2y/dx2=m2e^mx
Now substitute this and solve the equation......
y11 - 3y1/x = x d/dx (y1) - 3y1/x = x ..................1 put y1 = z theneq 1 becomes dz/dx - 3z/x = x ..............2 now this is a linear differential equation withintegration factor = 1/x3 z/x3 = x/x3 dx z/x3 = (1/x2)dx = -1/x + c1 z = -x2 + c1x3 ...................3 now , dy/dx = z so dy/dx = -x2 + c1x3 dy = (-x2 + c1x3)dx y = -x3/3 + c1x4/4 + c2 .................5 this is the required equation approve if u like my ans
y11 - 3y1/x = x
d/dx (y1) - 3y1/x = x ..................1
put y1 = z theneq 1 becomes
dz/dx - 3z/x = x ..............2
now this is a linear differential equation withintegration factor = 1/x3
z/x3 = x/x3 dx
z/x3 = (1/x2)dx = -1/x + c1
z = -x2 + c1x3 ...................3
now , dy/dx = z so
dy/dx = -x2 + c1x3
dy = (-x2 + c1x3)dx
y = -x3/3 + c1x4/4 + c2 .................5
this is the required equation
approve if u like my ans
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