Chapter 10

Equation of family of circles is \(\left(x^{2}+y^{2}-2 x-2 y+1\right)+\) $$ \lambda(3 x+4 y-7)=0 $$

The general equation of all non-horizontal lines in xy-plane is \(a x+b y=1\), where \(a \neq 0\)

\(\frac{1}{y+1} d y=-\frac{\cos x}{2+\sin x} d x\)

\((y \cos y+\sin y) d y=(2 x \log x+x) d x\)

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

\(d y-\sin x \sin y d x=0\)

We have \(\frac{d y}{d x}=\sin \frac{x-y}{2}-\sin \frac{x+y}{2}\) \(=-2 \cos \frac{x}{2} \sin \frac{y}{2}\)

Put \(x y=v\), i.e., \(y+x \frac{d y}{d x}=\frac{d v}{d x}\)

\(\frac{y^{\prime \prime \prime}}{y^{\prime \prime}}=8\) or \(\log y^{\prime \prime}=8 x+c\)

\(\frac{d y}{d x}=\frac{y}{x}\left[\log \frac{y}{x}+1\right]\)

\(\left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x\) or \(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}\)

The intersection of \(y-x+1=0\) and \(y+x+5=0\) is \((-2,-3)\). Put \(x=X-2, y=Y-3\).

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

\(x^{2} \frac{d y}{d x}-x y=1+\cos \frac{y}{x}\)

\(2 x^{2} y \frac{d y}{d x}=\tan \left(x^{2} y^{2}\right)-2 x y^{2}\)

\(\left\\{\frac{1}{x}-\frac{y^{2}}{(x-y)^{2}}\right\\} d x+\left\\{\frac{x^{2}}{(x-y)^{2}}-\frac{1}{y}\right\\} d y=0\)

The given equation can be rewritten as $$ \frac{d y}{d x}+\frac{x^{2}-1}{x\left(x^{2}+1\right)} y=\frac{x^{2} \log x}{\left(x^{2}+1\right)} $$

\(f^{\prime}(x)-\frac{2 x(x+1)}{x+1} f(x)=\frac{e^{x^{2}}}{(x+1)^{2}}\)

\(\frac{d x}{d y}=\frac{x+2 y^{3}}{y}\)

\(\frac{d y}{d x}=\frac{1}{x y\left[x^{2} \sin y^{2}+1\right]}\)

We have \(\frac{d y}{d x}=1-\frac{1}{x^{2}}\) or \(y=x+\frac{1}{x}+C\)