Question

Solve the simultaneous differential equations
dx÷dt+dy÷dt-2y=2cos t-7sin and dx÷dt-dy÷dt+2x=4cos t -3sin t​

Answer 1

To solve simultaneous differential equations of the form

$dx/dt + dy/dt - 2y = 2cos(t) - 7sin(t)$

and $dx/dt - dy/dt + 2x = 4cos(t) - 3sin(t),$

we can use an approach called the method of integrating factors.

The first step is to multiply both sides of each equation by an integrating factor (IF) that will make the left-hand side of each equation into the derivative of a product.

A common choice of IF for equations of this form is $e^(2t).$

$IF1 = e^(2t)$

$IF2 = e^(-2t)$

Multiply each side of the first equation by IF1:

$IF1 * dx/dt + IF1 * dy/dt - 2IF1 * y = 2IF1 * cos(t) - 7IF1 * sin(t)$

Now, take the derivative of both sides with respect to t:

$d/dt(IF1 * x) + d/dt(IF1 * y) - 2IF1 * dy/dt = 2IF1 * (-sin(t)) - 7IF1 * (-cos(t))$

The left hand side is the sum of the derivatives of two products, which equals the derivative of their sum.

$d/dt(IF1 * x + IF1 * y) = 2IF1 * sin(t) + 7IF1 * cos(t)$

Now, we can integrate both sides with respect to t:

$IF1 * x + IF1 * y = -2sin(t) + 7cos(t)$ + C

where C is a constant of integration

Similarly, we can solve for the second equation:

$d/dt(IF2 * x - IF2 * y) = 4IF2 * sin(t) + 3IF2 * cos(t)$

$IF2 * x - IF2 * y = 4sin(t) - 3cos(t) + D$

where D is a constant of integration

Solving for x and y, we get

$x = -2sin(t) + 7cos(t) + Ce^(2t)$

$y = 4sin(t) - 3cos(t) + De^(-2t)$

So, the general solution of the simultaneous differential equations is:

$x = -2sin(t) + 7cos(t) + Ce^(2t)$

$y = 4sin(t) - 3cos(t) + De^(-2t)$

where C and D are arbitrary constants, which can be determined by using the initial conditions or other given information.

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Answer 2

Step-by-step explanation:

To solve simultaneous differential equations of the form

dx/dt + dy/dt - 2y = 2cos(t) - 7sin(t)dx/dt+dy/dt−2y=2cos(t)−7sin(t)

and dx/dt - dy/dt + 2x = 4cos(t) - 3sin(t),dx/dt−dy/dt+2x=4cos(t)−3sin(t),

we can use an approach called the method of integrating factors.

The first step is to multiply both sides of each equation by an integrating factor (IF) that will make the left-hand side of each equation into the derivative of a product.

A common choice of IF for equations of this form is e^(2t).e

(

2t).

IF1 = e^(2t)IF1=e

(

2t)

IF2 = e^(-2t)IF2=e

(

−2t)

Multiply each side of the first equation by IF1:

IF1 * dx/dt + IF1 * dy/dt - 2IF1 * y = 2IF1 * cos(t) - 7IF1 * sin(t)IF1∗dx/dt+IF1∗dy/dt−2IF1∗y=2IF1∗cos(t)−7IF1∗sin(t)

Now, take the derivative of both sides with respect to t:

d/dt(IF1 * x) + d/dt(IF1 * y) - 2IF1 * dy/dt = 2IF1 * (-sin(t)) - 7IF1 * (-cos(t))d/dt(IF1∗x)+d/dt(IF1∗y)−2IF1∗dy/dt=2IF1∗(−sin(t))−7IF1∗(−cos(t))

The left hand side is the sum of the derivatives of two products, which equals the derivative of their sum.

d/dt(IF1 * x + IF1 * y) = 2IF1 * sin(t) + 7IF1 * cos(t)d/dt(IF1∗x+IF1∗y)=2IF1∗sin(t)+7IF1∗cos(t)

Now, we can integrate both sides with respect to t:

IF1 * x + IF1 * y = -2sin(t) + 7cos(t)IF1∗x+IF1∗y=−2sin(t)+7cos(t) + C

where C is a constant of integration

Similarly, we can solve for the second equation:

d/dt(IF2 * x - IF2 * y) = 4IF2 * sin(t) + 3IF2 * cos(t)d/dt(IF2∗x−IF2∗y)=4IF2∗sin(t)+3IF2∗cos(t)

IF2 * x - IF2 * y = 4sin(t) - 3cos(t) + DIF2∗x−IF2∗y=4sin(t)−3cos(t)+D

where D is a constant of integration

Solving for x and y, we get

x = -2sin(t) + 7cos(t) + Ce^(2t)x=−2sin(t)+7cos(t)+Ce

(

2t)

y = 4sin(t) - 3cos(t) + De^(-2t)y=4sin(t)−3cos(t)+De

(

−2t)

So, the general solution of the simultaneous differential equations is:

x = -2sin(t) + 7cos(t) + Ce^(2t)x=−2sin(t)+7cos(t)+Ce

(

2t)

y = 4sin(t) - 3cos(t) + De^(-2t)y=4sin(t)−3cos(t)+De

(

−2t)

where C and D are arbitrary constants, which can be determined by using the initial conditions or other given information.

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brainly.in/question/19810597

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