Solution for the question - the solution of the given differential equation (dy)/(dx)+2xy=y is y=ce^(-x^(2)) '/> y=ce^(x) '/> y=ce^(x^(2)-x) '/>

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

$y = c e^{x - x^{2}}$
$y = c e^{x^{2} - x}$
$y = c e^{x}$
$y = c e^{- x^{2}}$

hint Hint:

Use the separation of variables technique to solve the equation.

The correct answer is: $y equals c e to the power of x minus x to the power of 2 end exponent end exponent$

$fraction numerator d y over denominator d x end fraction space plus space 2 x y space equals space y fraction numerator d y over denominator d x end fraction space equals space y space minus space 2 x y fraction numerator d y over denominator d x end fraction space equals space y open parentheses 1 space minus space 2 x close parentheses fraction numerator d y over denominator y end fraction space equals space open parentheses 1 space minus space 2 x close parentheses d x I n t e g r a t i n g space b o t h space t h e space s i d e s comma integral fraction numerator d y over denominator y end fraction space equals space integral open parentheses 1 space minus space 2 x close parentheses d x log y space equals space open parentheses x space minus fraction numerator space 2 x squared over denominator 2 end fraction close parentheses space plus space c subscript 1 w h e r e space c subscript 1 space i s space a space c o n s tan t. log y space equals space x space minus space x squared space plus space c subscript 1 y space equals space e to the power of open parentheses x space minus x squared space plus space c subscript 1 close parentheses end exponent y space equals space space space e to the power of open parentheses x space minus x squared close parentheses end exponent space cross times space e to the power of c subscript 1 end exponent P u t t i n g space e to the power of c subscript 1 end exponent space equals space c comma therefore y space equals space c times e to the power of open parentheses x space minus space x squared close parentheses end exponent$

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 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

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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

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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

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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

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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

maths-General

General

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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

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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

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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

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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

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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

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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

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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

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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

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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

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The solution of the given differential equation is

Use the separation of variables technique to solve the equation.

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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 correct answer is: $y equals c e to the power of x minus x to the power of 2 end exponent end exponent$

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

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