Question

solve (y-x(dy/dx))=2y^2+dy/dx.
pls give step-by-step answer.​

Answer 1

EXPLANATION.

$\implies y - x \bigg(\dfrac{dy}{dx} \bigg) = 2y^{2} + \dfrac{dy}{dx}$

As we know that,

We can write equation as,

$\implies y = 2y^{2} + \dfrac{dy}{dx} + x \bigg(\dfrac{dy}{dx} \bigg)$

$\implies y - 2y^{2} = (1 + x) \dfrac{dy}{dx}$

$\implies (y - 2y^{2} ) dx = (1 + x)dy$

$\implies \dfrac{dx}{(1 + x)} = \dfrac{dy}{(y - 2y^{2}) }$

Integrate both sides of the equation, we get.

$\implies \displaystyle \int \dfrac{dx}{(1 + x)} = \int \dfrac{dy}{( y - 2y^{2}) }$

$\implies \displaystyle \int \dfrac{dx}{(1 + x)} \ = ln(1 + x)$

$\implies \displaystyle \int \dfrac{dy}{(y - 2y^{2} )} \ = \int \dfrac{dy}{y(1 - 2y)}$

As we know that,

Partial fractions is apply only when coefficient of denominator > coefficient of numerator.

Apply partial fractions in this equation, we get.

$\implies \displaystyle \dfrac{1}{y(1 - 2y)} \ = \dfrac{A}{y} \ + \dfrac{B}{(1 - 2y)}$

$\implies \displaystyle 1 = A(1 - 2y) + B(y)$

Put the value of y = 0 in the equation, we get.

$\implies \displaystyle 1 = A(1 - 2(0)) + B(0).$

$\implies \displaystyle 1 = A(1).$

$\implies \displaystyle A = 1. - - - - - (1).$

Put the value of y = 1/2 in the equation, we get.

$\implies \displaystyle 1 = A(1 - 2 \times (1/2)) + B(1/2)$

$\implies \displaystyle 1 = A(1 - 1) + B(1/2)$

$\implies \displaystyle 1 = B(1/2)$

$\implies \displaystyle B = 2. - - - - - (2).$

Put the value in the equation, we get.

$\implies \displaystyle \int \dfrac{A}{y} \ + \int \dfrac{B}{(1 - 2y)}$

$\implies \displaystyle \int \dfrac{1.dy}{y} \ + \int \dfrac{2.dy}{(1 - 2y)}$

$\implies \displaystyle ln(y) \ + \bigg(\dfrac{2}{-2} \bigg) ln(1 - 2y)$

$\implies \displaystyle ln(y) - ln(1 - 2y).$

Put the value in the main equation, we get.

$\implies \displaystyle \int \dfrac{dx}{(1 + x)} = \int \dfrac{dy}{( y - 2y^{2}) }$

$\implies \displaystyle ln( 1 + x) = ln(y) - ln( 1 - 2y) + C.$

$\implies \displaystyle ln(1 + x) = ln \bigg|\dfrac{y}{1 - 2y} \bigg| + C$