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Popular >

y^'=2t,y(0)=1

Solution

y^{^{'}} = 2 t, y (0) = 1

Solution

y = t^{2} + 1

Solution steps

y^{'} (t) = 2 t

If

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

then

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

y = \int 2 t d t

\int 2 t d t = t^{2} + c_{1}

y = t^{2} + c_{1}

Apply initial conditions:

y = t^{2} + 1

y = t^{2} + 1

Graph

Popular Examples

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

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y^{^{'}} = \frac{3 x ^{7} e ^{\frac{y}{x}} + x ^{5} y ^{2}}{x ^{6} y}

limit as x approaches 2 of (x-7)/(x+3)

x \to 2 lim (\frac{x - 7}{x + 3})

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\int_{1}^{\infty} \frac{1}{x ^{3} - 1} d x

integral of ysqrt(x+y^2)

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

Frequently Asked Questions (FAQ)

What is the solution for y^'=2t,y(0)=1 ?
The solution for y^'=2t,y(0)=1 is y=t^2+1