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Popular >

y^'=2x,y(1)=7

Solution

y^{^{'}} = 2 x, y (1) = 7

Solution

y = x^{2} + 6

Solution steps

y^{'} (x) = 2 x

If

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

then

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

y = \int 2 x d x

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

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

Apply initial conditions:

y = x^{2} + 6

y = x^{2} + 6

Popular Examples

xy^'=x^2sin(x)+y

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

tangent of f(x)=(36x)/(x^2+36)tangent of

f (x) = \frac{36 x}{x ^{2} + 36}

integral of ((4x^2))/((x-2)(x^2+4))

\int \frac{( 4 x ^{2} )}{( x - 2 ) ( x ^{2} + 4 )} d x

integral of (x^n)/n

\int \frac{x ^{n}}{n} d x

integral from 5 to 9 of ((x-4)/(x^2))

\int_{5}^{9} (\frac{x - 4}{x ^{2}}) d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=2x,y(1)=7 ?
The solution for y^'=2x,y(1)=7 is y=x^2+6