Question

General solution of \(y \frac{d y}{d x}+b y^{2}=a \cos x, 0<x<1\) is
(A) \(y^{2}=2 a(2 b \sin x+\cos x)+c e^{-2 b x}\)
(B) \(\left(4 b^{2}+1\right) y^{2}=2 a(\sin x+2 b \cos x)+c e^{-2 b x}\)
(C) \(\left(4 b^{2}+1\right) y^{2}=2 a(\sin x+2 b \cos x)+c e^{2 b x}\)
(D) \(y^{2}=2 a(2 b \sin x+\cos x)+c e^{-2 b x}\)
Here \(c\) is an arbitrary constant

Answer 1

Ans: (B)
Hint: Let \(y^{2}=z\)
$$
\begin{aligned}
&y \frac{d y}{d x}=\frac{1}{2} \frac{d z}{d x} \\
&\frac{d z}{d x}+2 b z=2 a \cos x \\
&I F=e^{2 b \int d x}=e^{2 b x} \\
&z \cdot e^{2 b x}=\int 2 a \cos x \cdot e^{2 b x} \cdot d x
\end{aligned}
$$

$$
\begin{aligned}
&y^{2} e^{2 b x}=\frac{2 a}{4 b^{2}+1}(\sin x+2 b \cos x) e^{2 b x}+c \\
&\left(4 b^{2}+1\right) y^{2}=2 a(\sin x+2 b \cos x)+c e^{-2 b x}
\end{aligned}
$$

Answer 2

order cialis 20mg sale <a href="https://ordergnonline.com/">tadalafil medication</a> herbal ed pills

Answer 3

canadian cialis online pharmacy <a href="https://ordergnonline.com/">cialis for sale online</a> ed pills for sale