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Popular >

(t^2+1)dx=(x+4)dt

Solution

(t^{2} + 1) d x = (x + 4) d t

Solution

t (x) = tan (ln (x + 4) - c_{1})

Solution steps

(t (x)^{2} + 1) d x = (x + 4) d t (x)

Solve separable ODE:

t (x) = tan (ln (x + 4) - c_{1})

t (x) = tan (ln (x + 4) - c_{1})

Popular Examples

2y(dy)/(dx)=xe^x

2 y \frac{d y}{d x} = x e^{x}

(dy)/(dx)y=arccsc(5x^2+1)

\frac{d y}{d x} y = arccsc (5 x^{2} + 1)

(dy}{dx}=\frac{10x)/y

\frac{d y}{d x} = \frac{10 x}{y}

(dy)/(dx)=2e^{x-y},y(0)=ln(8)

\frac{d y}{d x} = 2 e^{x - y}, y (0) = ln (8)

(x^2+1)(y^2-1)+xyy^'=0

(x^{2} + 1) (y^{2} - 1) + x y y^{^{'}} = 0

Frequently Asked Questions (FAQ)

What is the solution for (t^2+1)dx=(x+4)dt ?
The solution for (t^2+1)dx=(x+4)dt is t(x)=tan(ln(x+4)-c_{1})