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Popular >

y^'=e^{9x}-4x

Solution

y^{^{'}} = e^{9 x} - 4 x

Solution

y = \frac{1}{9} e^{9 x} - 2 x^{2} + c_{1}

Solution steps

y^{'} (x) = e^{9 x} - 4 x

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int e^{9 x} - 4 x d x

\int e^{9 x} - 4 x d x = \frac{1}{9} e^{9 x} - 2 x^{2} + c_{1}

y = \frac{1}{9} e^{9 x} - 2 x^{2} + c_{1}

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n = 1 \sum \infty 8 (0.2)^{n}

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n = 1 \sum \infty (1 + \frac{3}{n})^{n}

limit as x approaches-1 of (x^3-1)/(x-1)

x \to - 1 lim (\frac{x ^{3} - 1}{x - 1})

tangent of y=5x^3-5x,(1,0)tangent of

y = 5 x^{3} - 5 x, (1, 0)

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{9x}-4x ?
The solution for y^'=e^{9x}-4x is y= 1/9 e^{9x}-2x^2+c_{1}