# `y' + 2xy = 10x` Solve the first-order differential equation `y'+2xy =10x`

To solve, re-write the derivative as `dy/dx` .

`dy/dx + 2xy = 10x`

Then, bring together same variables on one side of the equation.

`dy/dx = 10x - 2xy`

`dy/dx = 2x(5 - y)`

`dy/(5-y) = 2x dx`

Next, take the integral of both sides.

`int dy/(5-y) =...

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`y'+2xy =10x`

To solve, re-write the derivative as `dy/dx` .

`dy/dx + 2xy = 10x`

Then, bring together same variables on one side of the equation.

`dy/dx = 10x - 2xy`

`dy/dx = 2x(5 - y)`

`dy/(5-y) = 2x dx`

Next, take the integral of both sides.

`int dy/(5-y) = int 2xdx`

`-ln |5-y| +C_1= (2x^2)/2 + C_2`

Then, isolate the y.

`-ln|5-y| = x^2+C_2-C_1`

`ln|5-y|=-x^2- C_2 +C_1`

Since C1 and C2 represent any number, express it as a single constant C.

`ln|5-y| = -x^2+ C`

`e^(ln|5-y|) = e^(-x^2+C)`

`|5-y| = e^(-x^2+C)`

`5-y = +-e^(-x^2+C)`

Applying the exponent rule `a^m*a^n = a^(m+n)` ,

the right side becomes

`5-y = +- e^(-x^2)*e^C`

`5-y = +-e^C*e^(-x^2)`

`-y = +-e^C*e^(-x^2) - 5`

`y = +-e^C*e^(-x^2)+5`

Since+-e^C is a constant, it can be replaced by a constant C.

`y = Ce^(-x^2) + 5`

Therefore, the general solution is  `y = Ce^(-x^2) + 5` .

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