Question

solve DE

2dy/dx+ 6 y/x = 2tan (x^4)

Answer 1

Question #7503 Solve $2dy/dx + 6y/x = 2\tan(x^4)$.

Solution. It is first order linear inhomogeneous equation. We are to use method, which is called variation of constants. First, look at the corresponding homogeneous equation: $2y' + 6y/x = 0$ or $y' = -3y/x$, which is equivalent to $dy/y = -3dx/x$, hence $\log |y| = -3\log |x| + C$, so $y = C \cdot x^{-3}$, where $C \in \mathbb{R}$. We must find the general solution of our equation in the form $y(x) = C(x) \cdot x^{-3}$, where $C(x)$ is a differentiable function. Substituting the last to our equation, one can get $2c'(x)x^{-3} - 6c(x)x^{-4} + 6c(x)x^{-4} = 2\tan x^4$ or $c'(x) = x^3\tan x^4$. Integrating the last $c(x) = \int x^3\tan x^4 dx = |z = x^4$, $dz = 4x^3 dx| = \frac{1}{4}\int \tan z dz = -1/4\log |\cos z| + C_1 = -1/4\log |\cos x^4| + C_1$, where $C_1$ is arbitrary real constant. Hence, the general solution of our equation is $y(x) = x^{-3} \cdot (-1/4\log |\cos x^4| + C_1)$, where $C_1 \in \mathbb{R}$.

Answer. $y(x) = x^{-3} \cdot (-1/4 \log |\cos x^4| + C_1)$, where $C_1 \in \mathbb{R}$.