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

解答

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

y = \frac{x ^{2} ln ( x )}{2} - \frac{x ^{2}}{4} + \frac{9}{4}

求解步骤

y^{'} (x) = x ln (x)

若

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

则

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

y = \int x ln (x) d x

\int x ln (x) d x = \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} + c_{1}

y = \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} + c_{1}

应用初始条件:

y = \frac{x ^{2} ln ( x )}{2} - \frac{x ^{2}}{4} + \frac{9}{4}

y = \frac{x ^{2} ln ( x )}{2} - \frac{x ^{2}}{4} + \frac{9}{4}

\int x (2 + x^{3})^{2} d x

\int I n^{2} d x

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

t an g e n t f (x) = 6 x^{2} - 6 x, at x = 0

\frac{d x}{d t} = (x^{2} - 1) t, x (0) = 0