The general solution of y frac{dy}{dx} + by^2 | vedclass

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(B) Given the differential equation: $y \frac{dy}{dx} + by^2 = a \cos x$. Let $y^2 = z$. Then, differentiating with respect to $x$, we get $2y \frac{d

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$(4b^2 + 1) y^2 = 2a(\sin x + 2b \cos x) + c e^{-2bx}$

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(B) Given the differential equation: $y \frac{dy}{dx} + by^2 = a \cos x$. Let $y^2 = z$. Then, differentiating with respect to $x$, we get $2y \frac{dy}{dx} = \frac{dz}{dx}$, which implies $y \frac{dy}{dx} = \frac{1}{2} \frac{dz}{dx}$. Substituting this into the original equation: $\frac{1}{2} \frac{dz}{dx} + bz = a \cos x$. Multiplying by $2$: $\frac{dz}{dx} + 2bz = 2a \cos x$. This is a linear differential equation of the form $\frac{dz}{dx} + Pz = Q$, where $P = 2b$ and $Q = 2a \cos x$. The integrating factor $(IF)$ is $e^{\int 2b \, dx} = e^{2bx}$. The solution is given by $z \cdot IF = \int Q \cdot IF \, dx + c$. $z e^{2bx} = \int 2a \cos x \cdot e^{2bx} \, dx + c$. Using the standard integral $\int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)$, we get: $z e^{2bx} = 2a \left[ \frac{e^{2bx}}{(2b)^2 + 1^2} (2b \cos x + \sin x) \right] + c$. $y^2 e^{2bx} = \frac{2a}{4b^2 + 1} e^{2bx} (2b \cos x + \sin x) + c$. Multiplying by $e^{-2bx}$: $y^2 = \frac{2a}{4b^2 + 1} (2b \cos x + \sin x) + c e^{-2bx}$. Rearranging gives: $(4b^2 + 1) y^2 = 2a(\sin x + 2b \cos x) + c e^{-2bx}$.

The general solution of $y \frac{dy}{dx} + by^2 = a \cos x$ for $0 \leq x < 1$ is (where $c$ is an arbitrary constant):

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