Math  /  Calculus

QuestionQ Q If y1=x^{y_{1}}=x is solution of (x2x)y+xy+y=0\left(x^{2}-x\right) y^{\prime \prime}+x y^{\prime}+y=0 find the General solution

Studdy Solution

STEP 1

1. The given differential equation is a second-order linear homogeneous differential equation.
2. The function y1 y_1 is a known solution of the differential equation.
3. We will use the method of reduction of order to find the general solution.

STEP 2

1. Verify that y1=x y_1 = x is a solution of the differential equation.
2. Use reduction of order to find a second, linearly independent solution.
3. Construct the general solution from the two linearly independent solutions.

STEP 3

Verify that y1=x y_1 = x is a solution of the differential equation by substituting it into the equation. Calculate the first and second derivatives of y1 y_1 :
y1=x y_1 = x y1=1 y_1' = 1 y1=0 y_1'' = 0
Substitute these into the differential equation:
(x2x)(0)+x(1)+x=0 (x^2 - x)(0) + x(1) + x = 0 x+x=0 x + x = 0 2x=0 2x = 0
This shows that y1=x y_1 = x is indeed a solution.

STEP 4

To find a second solution, use the method of reduction of order. Assume a solution of the form y2=v(x)y1=v(x)x y_2 = v(x) \cdot y_1 = v(x) \cdot x .
Calculate the derivatives:
y2=vx+v1 y_2' = v' \cdot x + v \cdot 1 y2=vx+2v y_2'' = v'' \cdot x + 2v'
Substitute y2,y2, y_2, y_2', and y2 y_2'' into the differential equation:
(x2x)(vx+2v)+x(vx+v)+vx=0 (x^2 - x)(v'' \cdot x + 2v') + x(v' \cdot x + v) + v \cdot x = 0
Simplify and solve for v(x) v(x) .

STEP 5

Simplify the equation from STEP_2:
(x3x2)v+(2x22x)v+x2v+xv=0 (x^3 - x^2)v'' + (2x^2 - 2x)v' + x^2v + xv = 0
Since y1=x y_1 = x is a solution, the terms involving v v should cancel out, leaving a simpler equation for v v' and v v'' . Assume v=0 v'' = 0 to simplify:
(2x22x)v=0 (2x^2 - 2x)v' = 0
This implies v=Cx2x v' = \frac{C}{x^2 - x} , where C C is a constant.
Integrate to find v(x) v(x) :
v(x)=Cx2xdx v(x) = \int \frac{C}{x^2 - x} \, dx

STEP 6

Perform the integration:
v(x)=Cx(x1)dx v(x) = \int \frac{C}{x(x-1)} \, dx
Use partial fraction decomposition:
1x(x1)=Ax+Bx1 \frac{1}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1}
Solve for A A and B B , then integrate:
v(x)=C(1xdx1x1dx) v(x) = C \left( \int \frac{1}{x} \, dx - \int \frac{1}{x-1} \, dx \right) v(x)=C(lnxlnx1) v(x) = C (\ln|x| - \ln|x-1|) v(x)=Clnxx1 v(x) = C \ln \left| \frac{x}{x-1} \right|

STEP 7

Construct the general solution using the two linearly independent solutions y1=x y_1 = x and y2=xlnxx1 y_2 = x \ln \left| \frac{x}{x-1} \right| :
y(x)=C1x+C2xlnxx1 y(x) = C_1 x + C_2 x \ln \left| \frac{x}{x-1} \right|
where C1 C_1 and C2 C_2 are arbitrary constants.
The general solution of the differential equation is:
y(x)=C1x+C2xlnxx1 y(x) = C_1 x + C_2 x \ln \left| \frac{x}{x-1} \right|

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