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y^{1/2}y^'+y^{3/2}=1,y(0)=4

解答

y^{\frac{1}{2}} y^{^{'}} + y^{\frac{3}{2}} = 1, y (0) = 4

解答

y = 3 49 e^{- 3 t} + 14 e^{- \frac{3 t}{2}} + 1

求解步骤

y^{\frac{1}{2}} y^{'} (t) + y^{\frac{3}{2}} = 1

求解

t, y^{'} (t) = \frac{d y}{d t} = \frac{1}{\frac{d t}{d y}} = \frac{1}{t ^{^{'}}}

y^{\frac{1}{2}} (\frac{1}{t ^{^{^{'}}}}) + y^{\frac{3}{2}} = 1

整理

t^{^{'}} : t^{^{'}} = - \frac{y ^{\frac{1}{2}}}{y ^{\frac{3}{2}} - 1}

若

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

则

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

t = \int - \frac{y ^{\frac{1}{2}}}{y ^{\frac{3}{2}} - 1} d y

\int - \frac{y ^{\frac{1}{2}}}{y ^{\frac{3}{2}} - 1} d y = - \frac{2}{3} ln (y^{\frac{3}{2}} - 1) + c_{1}

t = - \frac{2}{3} ln (y^{\frac{3}{2}} - 1) + c_{1}

应用初始条件:

t = - \frac{2}{3} ln (y^{\frac{3}{2}} - 1) + \frac{2 ln ( 7 )}{3}

整理

y : y = 3 49 e^{- 3 t} + 14 e^{- \frac{3 t}{2}} + 1

使用初始条件来验证解

y = 3 49 e^{- 3 t} + 14 e^{- \frac{3 t}{2}} + 1

作图

流行的例子

tangent y=x^4+7e^x,(0,7)

t an g e n t y = x^{4} + 7 e^{x}, (0, 7)

limit as x approaches 7+of 1/((x-7)^2)

x \to 7 + lim (\frac{1}{( x - 7 ) ^{2}})

integral of 1/(xsqrt(-1+x^2))

\int \frac{1}{x - 1 + x ^{2}} d x

derivative f(x)=-8

d er i v a t i v e f (x) = - 8

derivative \sqrt[7]{t}

d er i v a t i v e 7 t