Solution for the question - the general solution of y^(2)dx+(x^(2)-xy+y^(2))dy=0 is sinh^(-1)((x)/(y))+log y+c=0 '/> log(y+sqrt(x^(2)+y^(2)))+log y+c=0'/>

Maths-

General

Easy

Question

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

${t a n}^{- 1} (\frac{x}{y}) + l o g y + c = 0$
$2 {t a n}^{- 1} (\frac{x}{y}) + l o g x + c = 0$
$l o g (y + \sqrt{x^{2} + y^{2}}) + l o g y + c = 0$
${s i n h}^{- 1} (\frac{x}{y}) + l o g y + c = 0$

hint Hint:

Assume x/y as a trigonometric function in order to solve the question.

The correct answer is: $t a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator x over denominator y end fraction close parentheses plus l o g invisible function application y plus c equals 0$

$y squared d x space plus space open parentheses x squared minus x y plus y squared close parentheses d y space equals space 0 open parentheses x squared minus x y plus y squared close parentheses d y space equals space minus y squared d x fraction numerator open parentheses x squared minus x y plus y squared close parentheses over denominator negative y squared end fraction space equals space fraction numerator d x over denominator d y end fraction fraction numerator negative x squared over denominator y squared end fraction space plus space fraction numerator x y over denominator y squared end fraction space minus space y squared over y squared space equals space fraction numerator d x over denominator d y end fraction x over y space minus space open parentheses x over y close parentheses squared space minus space 1 space equals space fraction numerator d x over denominator d y end fraction space space space space space space space space space space........ open parentheses 1 close parentheses L e t space x over y space equals space tan theta space space space space space space space space space........ open parentheses 2 close parentheses therefore space x space equals space y tan theta D e r i v a t i n g space b o t h space t h e space s i d e s comma d x space equals space y S e c squared theta space d theta space plus space tan theta space d y space space space space space space space space....... open parentheses 3 close parentheses S u b s t i t u t i n g space e q n space open parentheses 2 close parentheses space a n d space open parentheses 3 close parentheses space i n space e q n space open parentheses 1 close parentheses comma space w e space g e t tan theta space minus space tan squared theta space minus 1 space equals fraction numerator y S e c squared theta space d theta space plus space tan theta space d y over denominator d y end fraction tan theta space minus space open parentheses 1 plus tan squared theta close parentheses space equals space fraction numerator y S e c squared theta space d theta over denominator d y end fraction space plus space fraction numerator tan theta space d y over denominator d y end fraction tan theta space minus space s e c squared theta space equals space space fraction numerator y S e c squared theta space d theta over denominator d y end fraction space plus space tan theta minus space s e c squared theta space equals space fraction numerator y S e c squared theta space d theta over denominator d y end fraction minus 1 space equals space fraction numerator y space d theta over denominator d y end fraction fraction numerator d y over denominator y end fraction space equals space minus 1 space d theta I n t e g r a t i n g space b o t h space t h e space s i d e s comma integral fraction numerator d y over denominator y end fraction space equals integral negative 1 space d theta log y space equals space theta space plus space c open parentheses F r o m space e q n space open parentheses 2 close parentheses comma theta space equals space tan to the power of negative 1 end exponent open parentheses x over y close parentheses close parentheses therefore log y space equals space tan to the power of negative 1 end exponent open parentheses x over y close parentheses space plus space c therefore tan to the power of negative 1 end exponent open parentheses x over y close parentheses space plus space log y space plus space c space equals space 0$

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

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

General

maths-

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

maths-General

General

maths-

Solution of differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y minus x over denominator y plus x end fraction$ is

maths-General

General

maths-

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

maths-General

General

maths-

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

maths-General

General

maths-

The solution of the differential equation $x d y minus y d x equals left parenthesis square root of x to the power of 2 end exponent plus y to the power of 2 end exponent right parenthesis end root d x$ is

maths-General

General

maths-

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

maths-General

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The general solution of is

Assume x/y as a trigonometric function in order to solve the question.

The correct answer is:

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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 correct answer is: $t a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator x over denominator y end fraction close parentheses plus l o g invisible function application y plus c equals 0$

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

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

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

Solution of differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y minus x over denominator y plus x end fraction$ is

Solution of differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y minus x over denominator y plus x end fraction$ is

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

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

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

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

The solution of the differential equation $x d y minus y d x equals left parenthesis square root of x to the power of 2 end exponent plus y to the power of 2 end exponent right parenthesis end root d x$ is

The solution of the differential equation $x d y minus y d x equals left parenthesis square root of x to the power of 2 end exponent plus y to the power of 2 end exponent right parenthesis end root d x$ is

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

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