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Popular >

y^'=(x+1)^2

Solution

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

Solution

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

Solution steps

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

If

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

then

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

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

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

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

Popular Examples

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

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

integral of sin(2)(pix)cos(5)(pix)

\int sin (2) (π x) cos (5) (π x) d x

y^{''}-2y^'+1y=0

y^{^{''}} - 2 y^{^{'}} + 1 y = 0

integral of (sin^3(θ))/(cos^6(θ))

\int \frac{sin ^{3} ( θ )}{cos ^{6} ( θ )} d θ

integral of (sin(2x))/2

\int \frac{sin ( 2 x )}{2} d x

Frequently Asked Questions (FAQ)

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