The resultant of two forces $P with not stretchy rightwards arrow on top$ and $Q with not stretchy rightwards arrow on top$ is of magnitude P. If the force is doubled, remaining unaltered, the new resultant will be

If the gradient of the tangent at any point (x, y) of a curve which passes through the point $open parentheses 1 comma fraction numerator pi over denominator 4 end fraction close parentheses$ is $open curly brackets fraction numerator y over denominator x end fraction minus s i n to the power of 2 end exponent invisible function application open parentheses fraction numerator y over denominator x end fraction close parentheses close curly brackets$ then equation of the curve is

If $y c o s invisible function application x plus x c o s invisible function application y equals pi$ , then $y to the power of ´ ´ end exponent left parenthesis 0 right parenthesis$ is

In this question, we have given ycosx + xcosy =π . Differentiate this equation with request to x and do separate second derivate and find the value of it.

If $y c o s invisible function application x plus x c o s invisible function application y equals pi$ , then $y to the power of ´ ´ end exponent left parenthesis 0 right parenthesis$ is

In this question, we have given ycosx + xcosy =π . Differentiate this equation with request to x and do separate second derivate and find the value of it.

The solution of the differential equation $x to the power of 4 end exponent fraction numerator d y over denominator d x end fraction plus x to the power of 3 end exponent y plus c o s e c invisible function application left parenthesis x y right parenthesis equals 0$ is equal to

In this question, you have to find the solution of differential equation x4 $fraction numerator d y over denominator d x end fraction$ + x3 y + cosec (xy) = 0. Use variable separable method to find the solution.

The solution of the differential equation $x to the power of 4 end exponent fraction numerator d y over denominator d x end fraction plus x to the power of 3 end exponent y plus c o s e c invisible function application left parenthesis x y right parenthesis equals 0$ is equal to

A continuously differentiable function $ϕ left parenthesis x right parenthesis in invisible function application left parenthesis 0 comma pi right parenthesis$ satisfying $y to the power of straight prime equals 1 plus y squared comma y left parenthesis 0 right parenthesis equals 0 equals y left parenthesis pi right parenthesis$ is

The slope of the tangent at $left parenthesis x comma y right parenthesis$ to a curve passing through $open parentheses 1 comma pi over 4 close parentheses$ is given by $y over x minus cos squared invisible function application open parentheses y over x close parentheses$ , then the equation of the curve is

A solution of the differential equation $open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared minus x fraction numerator d y over denominator d x end fraction plus y equals 0$ is

If integrating factor of $x open parentheses 1 minus x squared close parentheses d y plus open parentheses 2 x squared y minus y minus a x cubed close parentheses d x equals 0$ is $e to the power of integral P d x$ then P is equal to

The solution of the given differential equation $fraction numerator d y over denominator d x end fraction plus 2 x y equals y$ is

The general solution of $y squared d x plus open parentheses x squared minus x y plus y squared close parentheses d y equals 0$ is

The solution of the differential equation $fraction numerator d y over denominator d x end fraction equals y over x plus fraction numerator ϕ open parentheses y over x close parentheses over denominator ϕ to the power of straight prime open parentheses y over x close parentheses end fraction$ is

The solution of the differential equation $x space d y divided by d x equals y left parenthesis log invisible function application y minus log invisible function application x plus 1 right parenthesis space$ is

The solution of the differential equation $square root of a plus x end root fraction numerator d y over denominator d x end fraction plus x y equals 0$ is

A differential equation is an equation which contains one or more terms and the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable) dy/dx = f(x)

The solution of the differential equation $square root of a plus x end root fraction numerator d y over denominator d x end fraction plus x y equals 0$ is

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The solution of $fraction numerator d to the power of 2 end exponent y over denominator d x to the power of 2 end exponent end fraction equals sec to the power of 2 end exponent invisible function application x plus x e to the power of x end exponent$ is

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