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If y1 and y2 are two solutions of initial value problem y" + p(x)y' + q(x)y = 0, y(x0) = y0, y'(x0) = y0 and the Wronskion W(y1, y2) = 0, then y1 and y2 are
- a)linearly dependent
- b)linearly independent
- c)proportional
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If y1and y2are two solutions of initial value problem y" + p(x)y ...
If y1 and y2 are two solutions of an initial value problem y' = f(t, y) with initial condition y(t0) = y0, then their difference y = y1 - y2 satisfies the equation y' = f(t, y) with initial condition y(t0) = 0.
To see why, let's differentiate y = y1 - y2 with respect to t:
dy/dt = d(y1 - y2)/dt = dy1/dt - dy2/dt
Since y1 and y2 are solutions of the initial value problem, we have dy1/dt = f(t, y1) and dy2/dt = f(t, y2). Substituting these values into the equation above, we get:
dy/dt = f(t, y1) - f(t, y2)
But f(t, y1) = f(t, y2) since both y1 and y2 satisfy the equation y' = f(t, y). Therefore, dy/dt = 0, which means y is a constant.
To find the value of this constant, we use the initial condition y(t0) = y0. Since y1(t0) = y0 and y2(t0) = y0, we have y(t0) = y1(t0) - y2(t0) = y0 - y0 = 0.
Therefore, the difference y = y1 - y2 satisfies the equation y' = f(t, y) with initial condition y(t0) = 0.
To see why, let's differentiate y = y1 - y2 with respect to t:
dy/dt = d(y1 - y2)/dt = dy1/dt - dy2/dt
Since y1 and y2 are solutions of the initial value problem, we have dy1/dt = f(t, y1) and dy2/dt = f(t, y2). Substituting these values into the equation above, we get:
dy/dt = f(t, y1) - f(t, y2)
But f(t, y1) = f(t, y2) since both y1 and y2 satisfy the equation y' = f(t, y). Therefore, dy/dt = 0, which means y is a constant.
To find the value of this constant, we use the initial condition y(t0) = y0. Since y1(t0) = y0 and y2(t0) = y0, we have y(t0) = y1(t0) - y2(t0) = y0 - y0 = 0.
Therefore, the difference y = y1 - y2 satisfies the equation y' = f(t, y) with initial condition y(t0) = 0.
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If y1and y2are two solutions of initial value problem y" + p(x)y ...
Yes,wronskian becomes 0 then this is linearly dependent
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If y1and y2are two solutions of initial value problem y" + p(x)y + q(x)y = 0, y(x0) = y0, y(x0) = y0 and the Wronskion W(y1, y2) = 0, then y1and y2 area)linearly dependentb)linearly independentc)proportionald)None of theseCorrect answer is option 'B'. Can you explain this answer?
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