(y sin y+cos y +x) y-y

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Solution

Given function is y−cosy=x and corresponding differential function of the function is ( ysiny+cosy+x ) y ′ =y.

Differentiate both sides of given function with respect to x,

dy dx − d dx ( cosy )= d dx ( x ) y ′ −( −siny ) dy dx =1 y ′ +( siny ) y ′ =1 y ′ ( 1+siny )=1

Further simplify,

y ′ = 1 1+siny (1)

Substitute the value from equation (1) into L.H.S of given differential equation,

( ysiny+cosy+x ) y ′ =( ysiny+cosy+x )⋅ 1 1+siny

Since y−cosy=x, substitute the value of x,

( ysiny+cosy+x ) y ′ =( ysiny+cosy+y−cosy )⋅ 1 1+siny =( ysiny+y )⋅ 1 1+siny =y( 1+siny )⋅ 1 1+siny =y

That is equal to R.H.S.

Hence, it is verified that the given function is the solution of the corresponding differential equation

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Similar questions

For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

y=cosy=x and (ysiny+cosy+x)y'=y.

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10.

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y [x c o s (\frac{y}{x}) + y s i n (\frac{y}{x})] d x - x [y s i n (\frac{y}{x}) - x c o s (\frac{y}{x})] d y = 0

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