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Popular >

y^'=1+x^2

Solution

y^{^{'}} = 1 + x^{2}

Solution

y = x + \frac{x ^{3}}{3} + c_{1}

Solution steps

y^{'} (x) = 1 + x^{2}

If

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

then

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

y = \int 1 + x^{2} d x

\int 1 + x^{2} d x = x + \frac{x ^{3}}{3} + c_{1}

y = x + \frac{x ^{3}}{3} + c_{1}

Popular Examples

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\int \frac{16 x ^{2}}{( x - 6 ) ( x + 2 ) ^{2}} d x

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\frac{( k ^{2} x ^{3} - 6 k x + 9 )}{x - 1}

integral of x/(2x^2+12x+23)

\int \frac{x}{2 x ^{2} + 12 x + 23} d x

x(x+y)y^'=y(x-y)

x (x + y) y^{^{'}} = y (x - y)

Frequently Asked Questions (FAQ)

What is the solution for y^'=1+x^2 ?
The solution for y^'=1+x^2 is y=x+(x^3)/3+c_{1}