# 1) `y'+4x=2` , `y(0)=12` y(x) = ?? 2) `y''+18x=0` , `y(0)=5` , `y'(0)=10` y(x) = ?? and, find the slope of the tangent line to the polar curve for the given value of ...

1) `y'+4x=2` , `y(0)=12`

y(x) = ??

2) `y''+18x=0` , `y(0)=5` , `y'(0)=10`

y(x) = ??

and, find the slope of the tangent line to the polar curve for the given value of

`r=1/theta` , `theta=4`

m= ??

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The equation gives y' + 4x = 2 and it is known that y(0) = 12.

y' + 4x = 2

=> dy/dx = 2 - 4x

=> dy = 2 - 4x dx

Take the integral of both the sides

int dy = int 2 - 4x dx

=> y = 2x - 4x^2/2 + C

=> y = 2x - 2x^2 + C

As y(0) = 12

12 = 0 - 0 + C

**The function y = 2x - 2x^2 + 12**

The equation gives y'' + 18x = 0 and y(0) = 5, y'(0) = 10

y'' + 18x = 0

=> y'' = -18x

Take the integral of both the sides

y' = -18x^2/2 + C

=> y' = -9x^2 + C

As y'(0) = 10

10 = 0 + C

y' = -9x^2 + 10

Again take the integral

y = -9x^3/3 + 10x + C

=> y = -3x^3 + 10x + C

As y(0) = 5

C = 5

**The function y(x) = -3x^3 + 10x + 5**