Question Q) Solve the following: dx dx O dy dx + 2X 4 x² 1+ y² + 2+x² ។ 24- . @ cos x sinx dx _yx?dy tydy - xylxo 3 ③ d'y dy 2x² -4 ax +5y=0 dx 2 dx @x-y song sa citeste y ay ( xt dy Q) Solve the following: dx dx O dy dx + 2X 4 x² 1+ y² + 2+x² ។ 24- . @ cos x sinx dx _yx?dy tydy - xylxo 3 ③ d'y dy 2x² -4 ax +5y=0 dx 2 dx @x-y song sa citeste y ay ( xt dy

Solution-  Solution- (1) The given differential equation isd x d y+\frac{2 x}{1+x^{2}} y d x=\frac{4 x^{2}}{1+x^{2}} d xDividing by d x on both the sides, we have\frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{4 x^{2}}{1+x^{2}}Whose I.F is given by\text { I.F. }=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}Thus the solution is given byy x\left(1+x^{2}\right)=\int\left(1+x^{2}\right) x \frac{4 x^{2}}{1+x^{2}} d x+Cwhere C is an arbitrary constant.\boldsymbol{y} \boldsymbol{x}\left(1+x^{2}\right)=\frac{4 x^{3}}{3}+\boldsymbol{C}Hencey=\frac{4 x^{3}}{3 x\left(1+x^{2}\right)}+\frac{C}{x\left(1+x^{2}\right)}which is the required general solution.Solution- (2) - The given differential equation is\cos (x) \sin (x) d x-y x^{2} d y+y d y-x y^{2} d x=0Dividing by d x on both the sides, we have\begin{array}{l} \cos (x) \sin (x)-y x^{2} \frac{d y}{d x}+y \frac{d y}{d x}-x y^{2}=0 \\\Rightarrow \cos (x) \sin (x)-x y^{2}=y \frac{d y}{d x}\left(x^{2}-1\right)\end{array}Multiplying by 2 on both the sides, we have\sin (2 x)-2 x y^{2}=2 y \frac{d y}{d x}\left(x^{2}-1\right)Or2 y \frac{d y}{d x}+\frac{2 x y^{2}}{x^{2}-1}=\frac{\sin (2 x)}{x^{2}-1}Puttingy^{2}=z\Longrightarrow 2 y \frac{d y}{d x}=\frac{d z}{d x} \Longrightarrow \frac{d z}{d x}+\frac{2 x z}{x^{2}-1}=\frac{\sin (2 x)}{x^{2}-1}Whose I.F. is given byI.F. =e^{\int \frac{2 x}{x^{2}-1} d x}=e^{\log \left(x^{2}-1\right)}=x^{2}-1Thus the solution is given byz \times\left(x^{2}-1\right)=\int\left(x^{2}-1\right) \times \frac{\sin (2 x)}{x^{2}-1} d x+Cwhere C is an arbitrary constant .Substituting z=y^{2}, theny^{2}\left(x^{2}-1\right)=\frac{-1}{2} \cos (2 x)+CHencey^{2}=\frac{-1}{2\left(x^{2}-1\right)} \cos (2 x)+\frac{C}{x^{2}-1}which is the required result .Solution - (3) The given differential equation is\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+5 y=0The auxiliary equation ism^{2}-4 m+5=0By using the quadratic formulam=\frac{4 \pm \sqrt{16-20}}{2}=\frac{4 \pm 2 i}{2}=2 \pm iThus the general solution is given by\boldsymbol{y}=e^{2 x}\left(c_{1} \cos (x)+c_{2} \sin (x)\right)Required solution.Solution - (4) The given differential equation is\boldsymbol{x}-\boldsymbol{y} \frac{d x}{d y}=a\left(x^{2}+\frac{d x}{d y}\right)OR\boldsymbol{x}-\boldsymbol{y} \frac{d x}{d y}=a x^{2}+a \frac{d x}{d y}\Longrightarrow x(1-a x)=(y+a) \frac{d x}{d y}Separating the variables, we have\frac{d y}{y+a}=\frac{d x}{x(1-a x)}Or\frac{d y}{y+a}=\frac{d x}{x}+\frac{a \cdot d x}{(1-a x)}Integrating on both the sides, we have\int \frac{d y}{y+a}=\int \frac{d x}{x}+\int \frac{a \cdot d x}{(1-a x)}+\log (C)where \log (C) is an arbitrary constant.\begin{array}{l}\log (y+a)=\log (x)-\log (1-a x)+\log (C) \\\Longrightarrow \log (y+a)=\log \left(\frac{C x}{1-a x}\right)\end{array}OR\boldsymbol{y}+\boldsymbol{a}=\left(\frac{C x}{1-a x}\right)Hence\boldsymbol{y}=\left(\frac{C x}{1-a x}\right)-aRequired solution. ...