If two forces $P plus Q$ and $P minus Q$ make an angle $2 alpha$ with each other and their resultant makes an angle $theta$ with the bisector of the angle between the two forces, then is equal to

The maximum resultant of two forces is P and the minimum resultant is $Q$ , the two forces are at right angles, the resultant is

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

The resultant of two forces 3P and 2P is R, if the first force is doubled, the resultant is also doubled. The angle between the forces is

In this question, we have to make two equations of Resultant as per given instruction and solve the equations and find the angle. Remember the formula of Resultant is R = $square root of left parenthesis a squared plus b squared plus 2 a b space cos theta right parenthesis end root.$

The resultant of two forces 3P and 2P is R, if the first force is doubled, the resultant is also doubled. The angle between the forces is

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