What is the solution for (x+2)dt=2t+3dx ?

The solution for (x+2)dt=2t+3dx is t(x)=(e^{2c_{1}}x^2)/2+2e^{2c_{1}}x+2e^{2c_{1}}-3/2

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(x+2)dt=2t+3dx

(x + 2) d t = 2 t + 3 d x

t (x) = \frac{e ^{2 c_{1}} x ^{2}}{2} + 2 e^{2 c_{1}} x + 2 e^{2 c_{1}} - \frac{3}{2}

Solution steps

(x + 2) d t (x) = 2 t (x) + 3 d x

Solve separable ODE:

t (x) = \frac{e ^{2 c_{1}} x ^{2}}{2} + 2 e^{2 c_{1}} x + 2 e^{2 c_{1}} - \frac{3}{2}

t (x) = \frac{e ^{2 c_{1}} x ^{2}}{2} + 2 e^{2 c_{1}} x + 2 e^{2 c_{1}} - \frac{3}{2}

\frac{\partial}{\partial x} (\frac{x ^{2}}{4})

\frac{\partial}{\partial x} (x^{6} y^{4} - x^{5} y^{3})

\int (\frac{1}{x ^{2} - 4}) d x

(0, - 3), (- 4, 2)

s^{3}

What is the solution for (x+2)dt=2t+3dx ?
The solution for (x+2)dt=2t+3dx is t(x)=(e^{2c_{1}}x^2)/2+2e^{2c_{1}}x+2e^{2c_{1}}-3/2