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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

$2 c o s (x y) + x^{- 2} = c$
$2 c o s (x y) + y^{- 2} = c$
$2 s i n (x y) + x^{- 2} = c$
$2 s i n (x y) + y^{- 2} = c$

hint Hint:

In this question, we 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 to find the solution.

The correct answer is: $2 c o s invisible function application left parenthesis x y right parenthesis plus x to the power of negative 2 end exponent equals c$

Integrating both sides,

; .

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.

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

Maths-General

General

Maths-

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

Maths-General

General

Maths-

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

Maths-General

General

Maths-

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

Maths-General

General

Maths-

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

Maths-General

General

Maths-

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

Maths-General

General

Maths-

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

Maths-General

General

Maths-

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

Maths-General

General

Maths-

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

Maths-General

General

maths-

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

maths-General

General

maths-

The solution of the differential equation $x fraction numerator d to the power of 2 end exponent y over denominator d x to the power of 2 end exponent end fraction equals 1$ , given that $y equals 1 comma fraction numerator d y over denominator d x end fraction equals 0$ when $x equals 1$ , is

maths-General

General

maths-

Solution of $cos invisible function application x fraction numerator d y over denominator d x end fraction plus y sin invisible function application x equals 1$ is

maths-General

General

maths-

Integrating factor of $fraction numerator d y over denominator d x end fraction plus fraction numerator y over denominator x end fraction equals x to the power of 3 end exponent minus 3$ is

maths-General

General

maths-

Integrating factor of differential equation $cos invisible function application x fraction numerator d y over denominator d x end fraction plus y sin invisible function application x equals 1$ is

maths-General

General

maths-

The solution of the differential equation $fraction numerator d y over denominator d x end fraction plus y equals cos invisible function application x$ is

maths-General

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The solution of the differential equation is equal to

In this question, we have to find the solution of differential equation x4 + x3 y + cosec (xy) = 0. Use variable separable to find the solution.

The correct answer is:

Integrating both sides, ; .

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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, we 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 to find the solution.

The correct answer is: $2 c o s invisible function application left parenthesis x y right parenthesis plus x to the power of negative 2 end exponent equals c$

Integrating both sides,

; .

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

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

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 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 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 $fraction numerator d y over denominator d x end fraction equals y over x plus fraction numerator ϕ open parentheses y over x close parentheses over denominator ϕ to the power of straight prime open parentheses y over x close parentheses end fraction$ 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 $x space d y divided by d x equals y left parenthesis log invisible function application y minus log invisible function application x plus 1 right parenthesis space$ 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

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

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

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

The solution of the differential equation $x fraction numerator d to the power of 2 end exponent y over denominator d x to the power of 2 end exponent end fraction equals 1$ , given that $y equals 1 comma fraction numerator d y over denominator d x end fraction equals 0$ when $x equals 1$ , is

The solution of the differential equation $x fraction numerator d to the power of 2 end exponent y over denominator d x to the power of 2 end exponent end fraction equals 1$ , given that $y equals 1 comma fraction numerator d y over denominator d x end fraction equals 0$ when $x equals 1$ , is

Solution of $cos invisible function application x fraction numerator d y over denominator d x end fraction plus y sin invisible function application x equals 1$ is

Solution of $cos invisible function application x fraction numerator d y over denominator d x end fraction plus y sin invisible function application x equals 1$ is

Integrating factor of $fraction numerator d y over denominator d x end fraction plus fraction numerator y over denominator x end fraction equals x to the power of 3 end exponent minus 3$ is

Integrating factor of $fraction numerator d y over denominator d x end fraction plus fraction numerator y over denominator x end fraction equals x to the power of 3 end exponent minus 3$ is

Integrating factor of differential equation $cos invisible function application x fraction numerator d y over denominator d x end fraction plus y sin invisible function application x equals 1$ is

Integrating factor of differential equation $cos invisible function application x fraction numerator d y over denominator d x end fraction plus y sin invisible function application x equals 1$ is

The solution of the differential equation $fraction numerator d y over denominator d x end fraction plus y equals cos invisible function application x$ is

The solution of the differential equation $fraction numerator d y over denominator d x end fraction plus y equals cos invisible function application x$ is