Question

If dx/dt +5y = 0 and dy/dt - 5x = 0
then y​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ \frac{dx}{dt} + 5y = 0}$

$\displaystyle \sf{ \frac{dy}{dt} + 5x = 0}$

TO DETERMINE

The value of y

EVALUATION

Here the given system of differential equations are

$\displaystyle \sf{ \frac{dx}{dt} + 5y = 0} \: \: \: \: \: - - - - (1)$

$\displaystyle \sf{ \frac{dy}{dt} + 5x= 0} \: \: \: - - - - (2)$

Differentiating both sides of Equation 2 we get

$\displaystyle \sf{ \frac{ {d}^{2} y}{d {t}^{2} } + 5 \frac{dx}{dt} = 0}$

$\displaystyle \sf{ \implies \: \frac{ {d}^{2} y}{d {t}^{2} } = - 5 \frac{dx}{dt} }$

From Equation 1 we get

$\displaystyle \sf{ \implies \: \frac{ {d}^{2} y}{d {t}^{2} } = - 5 \times ( - 5y)}$

$\displaystyle \sf{ \implies \: \frac{ {d}^{2} y}{d {t}^{2} } - 25y = 0}$

$\displaystyle \sf{ \implies ({D }^{2} - 25)y = 0 }$

$\displaystyle \sf{ Let \: \: y = {e}^{mx} \: \: be \: the \: trial \: solution }$

Then the auxiliary equation is

$\displaystyle \sf{ {m }^{2} - 25 = 0 }$

$\displaystyle \sf{ \implies{m }^{2} = 25 }$

$\displaystyle \sf{ \implies \: m = \pm \: 5}$

Thus we have

$\displaystyle \sf{ y = A {e}^{5t} +B {e}^{ - 5t} }$

From Equation 2 we get

$\displaystyle \sf{5x = - \frac{dy}{dt} }$

$\displaystyle \sf{ \implies \: 5x = - 5 A {e}^{5t} + 5 B {e}^{ - 5t} }$

$\displaystyle \sf{ \implies \: x = - A {e}^{5t} + B {e}^{ - 5t} }$

Thus we get

$\displaystyle \sf{ x = - A {e}^{5t} + B {e}^{ - 5t} }$

$\displaystyle \sf{ y= A {e}^{5t} + B {e}^{ - 5t} }$

FINAL ANSWER

$\displaystyle \sf{ y= A {e}^{5t} + B {e}^{ - 5t} }$

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