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Popular >

y^'=x^2+1

Solution

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

Solution

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

Solution steps

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

If

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

then

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

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

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

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

Popular Examples

integral of t/((t^2+1)^{15/2)}

\int \frac{t}{( t ^{2} + 1 ) ^{\frac{15}{2}}} d t

integral of (x^6-2x^3)^5(x^5-x^2)

\int (x^{6} - 2 x^{3})^{5} (x^{5} - x^{2}) d x

derivative of 1/(\sqrt[3]{(8-x^3)^8)}derivative of

area x=y^2-4y,x=-y^2+4y area

x = y^{2} - 4 y, x = - y^{2} + 4 y

tangent of y=x^2-x,(4,12)tangent of

y = x^{2} - x, (4, 12)

Frequently Asked Questions (FAQ)

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