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Popular >

t^2y^'+ty=7,y(1)=4

Solution

t^{2} y^{^{'}} + t y = 7, y (1) = 4

Solution

y = \frac{7 ln ( t )}{t} + \frac{4}{t}

Solution steps

t^{2} y^{'} (t) + t y = 7

Solve linear ODE:

y = \frac{7 ln ( t )}{t} + \frac{4}{t}

y = \frac{7 ln ( t )}{t} + \frac{4}{t}

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

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x \to \infty lim (\frac{0}{1})

(\partial)/(\partial x)(((9y+7z))/(x+u))

\frac{\partial}{\partial x} (\frac{( 9 y + 7 z )}{x + u})

integral of (3r^2)/(r+5)

\int \frac{3 r ^{2}}{r + 5} d r

slope ofintercept (3,5),(1,-7)slope intercept

(3, 5), (1, - 7)

Frequently Asked Questions (FAQ)

What is the solution for t^2y^'+ty=7,y(1)=4 ?
The solution for t^2y^'+ty=7,y(1)=4 is y=(7ln(t))/t+4/t