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# Answer to Question #106446 in Mechanics | Relativity for Najeya

Question #106446
Solve dy/dt + ycotx = e^cosx for x= pi by two and y= -2
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Expert's answer
2020-03-25T10:59:04-0400

### Mechanics | Relativity

We need to solve the given Differential equation

Solution:

Given equation is the Linear differential in the form "\\frac {dy}{dx} + P(x) y = Q(x)"

Now compare the given equation "\\frac {dy}{dx} + y \\space cot x = e^ {cos \\space x}" with the above form

"P(x) = cot \\space x"

"Q(x) = e^{cos \\space x}"

Integrating factor = I.F = "e^{\\int P(x) dx} = e^ {\\int cot x \\space dx} = e^ {ln \\space sin x} = sin \\space x"

The solution of the Linear equation is,

"y \\times (I.F) = \\int (I.F) \\times Q(x) dx + c"

"y \\times sin x = \\int sin x \\space e^{cos x} dx + c"

"y \\space sin x = - \\int d(cos x) \\space e^{cos x} dx + c"

"y \\space sin x = - e^{cos x} + c"

Now we need to find the c value by plugging "x = \\frac {\\pi}{2} \\space and \\space y =-2"

"-2 (sin \\frac{\\pi}{2} ) = - e^{cos {\\frac {\\pi}{2 }}} + c"

"-2 =- e^0 + c \\\\\n-2 = -1 +c \\\\\nc = -1"

Finally the solution is ,

"y \\space sin x = - e^{cos x} -1"

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