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Solving Linear Differential Equations of First Order

Solve: 2x-1...

Question

Solve:
$(2 x - 10 y^{3}) d y + y d x = 0, y \neq 0$

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Solution

Simplifying $(2 x - 10 y^{3}) d y + y d x = 0$

$(2 x - 10 y^{3}) d y = - y d x$

$- y \frac{d x}{d y} = 2 x - 10 y^{3}$

$y \frac{d x}{d y} = 10 y^{3} - 2 x$

$y \frac{d x}{d y} + 2 x = 10 y^{3}$

$\frac{d x}{d y} + 2 \frac{x}{y} = 10 y^{2}$

$I . F . = e^{\int \frac{2}{y} d y}$

$= e^{2 log y}$

$= e^{log y^{2}}$

$= y^{2}$

Now,

$x (I . F .) = \int 10 y^{2} (I . F .) d y$

$x (y^{2}) = 10 \int y^{2} (y^{2}) d y$

$x y^{2} = 10 \int y^{4} d y$

$x y^{2} = 10 \frac{y^{5}}{5} + c$

$x y^{2} = 2 y^{5} + c$

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Similar questions

Q. The solution of the differential equation

(2 x + 10 y^{3}) \frac{d y}{d x} + y = 0

Q. The solution of the differential equation:

(2 x - 10 y^{3}) \frac{d y}{d x} + y = 0

Q. If the solution curve of the differential equation

(2 x - 10 y^{3}) d y + y d x = 0,

passes through the points

(0, 1)

(2, β)

β

is a root of the equation

y d x + (x - y^{2}) d y = 0

2 y e^{\frac{x}{y}} d x + ⎛ ⎜ ⎝ y - 2 x e^{\frac{x}{y}} ⎞ ⎟ ⎠ d y = 0

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