Dy Dx Cot Y X Cos Y Cot Y 0

Here we will use. Let y vx solving sin putting y vx diff w r t x x v x v putting value of 2 2 2 and y vx in 1 y x sin v sin v sin v v sin v v sin sin integrating both sides log cot log log log cot log log log cot log x cosec v cot v c x 1 sin cos sin c x 1 cos sin c x 1 cos v c sin v. Y y cot x.

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Example 22 Particular Solution Dy Dx Y Cot X 2x X2 Cot X

Misc 16 If Cos Y X Cos A Y Prove Dy Dx Cos2 A Y

$Solution Solve Cosxcosy Cotx Dx Sinxsinydy 0$

Solution Solve Cosxcosy Cotx Dx Sinxsinydy 0

Ex 9 6 9 Find General Solution X Dy Dx Y X Xy Cot X

Ex 9 6 15 Find Particular Solution Dy Dx 3y Cot X Sin 2x

The standard approach is to find a solution y c of the homogeneous equation by looking at the auxiliary equation which is the quadratic equation with the coefficients of the derivatives and then finding an independent particular solution y p of the non homogeneous equation.

Dy dx cot y x cos y cot y 0. So the differential equation is homogenous step 3. We shall use both implicit differentiation and chain rule. Dy dx y x i am assuming that you want to find dy dx. Free math problem solver answers your algebra geometry trigonometry calculus and statistics homework questions with step by step explanations just like a math tutor.

D dx uv udv dx v du dx. Ex 9 6 9 for each of the differential equation find the general solution. Xy cot xy differentiate both sides with respect to x to get. Cot y dy cot x dx cos y dy sin y cos x dx sin x d sin y sin y d sin x sin x ln sin y ln sin x ln c sin y csin x sin x 0.

A this is a second order linear non homogeneous differentiation equation. D dx xy d dx cot xy apply chain rule d dx xy csc 2 xy d dx xy 1 csc 2 xy d dx xy 0 d dx xy 0 since 1 csc 2 xy is always positive apply chain rule again y x dy dx 0 dy dx. X cot y.

Cot x cos x sin x cosec 2 x 1 cot 2 x. Dy dt dt dx dy dx.