If x siny-x d y-y siny-x-x- d x- x-0 and y1-2 then the value of cosy-x is

The correct option is B log xsin ( yx ) dydx = yxsinyx 1 ⇒dydx = yxsinyx 1/sinyx Let yx = t ⇒dydx = t+x dtdx ⇒ t+x dtdx= t 1sin t⇒ ∫sin t dt = ∫dxx ⇒cos ...

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Byju's Answer

Standard XII

Mathematics

Property 1

If x siny/x d...

Question

If $x sin (\frac{y}{x}) d y = [y sin (\frac{y}{x}) - x] d x, x > 0 and y (1) = \frac{π}{2}$ then the value of $cos (\frac{y}{x})$ is

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Solution

The correct option is B $log x$
$sin (\frac{y}{x}) \frac{d y}{d x} = \frac{y}{x} sin \frac{y}{x} - 1 \Rightarrow \frac{d y}{d x} = \frac{\frac{y}{x} sin \frac{y}{x} - 1}{sin \frac{y}{x}} Let \frac{y}{x} = t \Rightarrow \frac{d y}{d x} = t + x \frac{d t}{d x} \Rightarrow t + x \frac{d t}{d x} = t - \frac{1}{sin t} \Rightarrow - \int sin t d t = \int \frac{d x}{x} \Rightarrow cos t = log x + c \Rightarrow cos \frac{y}{x} = log x + c y (1) = \frac{π}{2} \Rightarrow cos \frac{π}{2} = log 1 + c \Rightarrow c = 0 \Rightarrow cos (\frac{y}{x}) = log x$

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Similar questions

Q. The solution of

[x + y sin (\frac{y}{x})] d x = x sin (\frac{y}{x}) d y

is:

Q. Solve:

(x sin \frac{y}{x}) d y = (y sin \frac{y}{x} - x) d x

Q. Solve the differential equation:

x sin (\frac{y}{x}) d y = (y sin (\frac{y}{x}) - x) d x

Q. In the following, show that the given differential equation is homogeneous and solve each of them.
1.

(x^{2} + x y) d y = (x^{2} + y^{2}) d x

y^{'} = \frac{x + y}{x}

(x y) d y (x + y) d x = 0

(x^{2} + y^{2}) d x + 2 x y d y = 0

x^{2} \frac{d y}{d x} = x^{2} - 2 y^{2} + x y

x d y . y d x = \sqrt{x^{2} + y^{2}} d x

{x cos (\frac{y}{x}) + y sin (\frac{y}{x})} y d x = {y sin (\frac{y}{x}) - x cos (\frac{y}{x})} x d y

x \frac{d y}{d x} - y + x sin (\frac{y}{x}) = 0

y d x + x log (\frac{y}{x}) d y - 2 x d y = 0

Q. Find the particular solution of the differential equation:

x \frac{d y}{d x} - y + x sin (\frac{y}{x}) = 0

, given that when

x = 2, y = π .

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If xsin(yx)dy=[ysin(yx)−x]dx,x>0 and y(1)=π2 then the value of cos(yx) is

The correct option is B logxsin(yx)dydx=yxsinyx−1⇒dydx=yxsinyx−1sinyxLet yx=t⇒dydx=t+xdtdx⇒t+xdtdx=t−1sin t⇒−∫sin t dt=∫dxx⇒cos t=logx+c⇒cos yx=logx+cy(1)=π2⇒cosπ2=log1+c⇒ c=0⇒cos(yx)=log x

If $x sin (\frac{y}{x}) d y = [y sin (\frac{y}{x}) - x] d x, x > 0 and y (1) = \frac{π}{2}$ then the value of $cos (\frac{y}{x})$ is