\displaystyle \text{Verify that the given function is a solution of the given differential equation:}

\displaystyle \textbf{Question 1: }~\text{Show that }y=be^{x}+ce^{2x}\text{ is a solution of the differential equation } \\ \frac{d^{2}y}{dx^{2}}-3\frac{dy}{dx}+2y=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = b e^{x} + c e^{2x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = b e^{x} + 2c e^{2x} \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = b e^{x} + 4c e^{2x}
\displaystyle = 3b e^{x} + 6c e^{2x} - 2b e^{x} - 2c e^{2x}
\displaystyle = 3\left(b e^{x} + 2c e^{2x}\right) - 2\left(b e^{x} + c e^{2x}\right)
\displaystyle = 3\frac{dy}{dx} - 2y \qquad \text{[Using equations (1) and (2)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} - 3\frac{dy}{dx} + 2y = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 2: }~\text{Verify that }y=4\sin 3x\text{ is a solution of the differential equation } \\ \frac{d^{2}y}{dx^{2}}+9y=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have, } y = 4\sin 3x \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = 12\cos 3x \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = -36\sin 3x
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -9(4\sin 3x)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -9y \qquad \text{[Using equation (1)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} + 9y = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 3: }~\text{Show that }y=ae^{2x}+be^{-x}\text{ is a solution of the differential equation } \\ \frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}-2y=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = a e^{2x} + b e^{-x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = 2a e^{2x} - b e^{-x} \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = 4a e^{2x} + b e^{-x}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 2a e^{2x} - b e^{-x} + 2a e^{2x} + 2b e^{-x}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = (2a e^{2x} - b e^{-x}) + 2(a e^{2x} + b e^{-x})
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + 2y \qquad \text{[Using equations (1) and (2)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} - \frac{dy}{dx} - 2y = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 4: }~\text{Show that the function }y=A\cos x+B\sin x\text{ is a solution of the differential equation } \\ \frac{d^{2}y}{dx^{2}}+y=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = A\cos x + B\sin x \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = -A\sin x + B\cos x \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = -A\cos x - B\sin x
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -(A\cos x + B\sin x)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -y \qquad \text{[Using equation (1)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} + y = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 5: }~\text{Show that the function }y=A\cos 2x-B\sin 2x\text{ is a solution of the differential equation } \\ \frac{d^{2}y}{dx^{2}}+4y=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = A\cos 2x - B\sin 2x \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = -2A\sin 2x - 2B\cos 2x \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = -4A\cos 2x + 4B\sin 2x
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -4(A\cos 2x - B\sin 2x)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -4y \qquad \text{[Using equation (1)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} + 4y = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 6: }~\text{Show that }y=Ae^{Bx}\text{ is a solution of the differential equation } \\ \frac{d^{2}y}{dx^{2}}=\frac{1}{y}\left(\frac{dy}{dx}\right)^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = A e^{Bx} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = AB e^{Bx} \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = AB^{2} e^{Bx}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{(AB e^{Bx})^{2}}{A e^{Bx}}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{1}{y}\left(\frac{dy}{dx}\right)^{2} \qquad \text{[Using equations (1) and (2)]}
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 7: }~\text{Verify that }y=\frac{a}{x}+b\text{ is a solution of the differential equation } \\ \frac{d^{2}y}{dx^{2}}+\frac{2}{x}\left(\frac{dy}{dx}\right)=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = \frac{a}{x} + b \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = -\frac{a}{x^{2}} \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{2a}{x^{3}}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{2}{x}\left(-\frac{a}{x^{2}}\right)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = -\frac{2}{x}\left(\frac{dy}{dx}\right) \qquad \text{[Using equation (2)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} + \frac{2}{x}\left(\frac{dy}{dx}\right) = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 8: }~\text{Verify that }y^{2}=4ax\text{ is a solution of the differential equation } \\ y=x\frac{dy}{dx}+a\frac{dx}{dy}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y^{2} = 4ax \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle 2y\frac{dy}{dx} = 4a
\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2a}{y} \qquad (2)
\displaystyle \text{Now, differentiating both sides of equation (1) with respect to } y, \text{ we get}
\displaystyle 2y = 4a\frac{dx}{dy}
\displaystyle \Rightarrow \frac{dx}{dy} = \frac{y}{2a} \qquad (3)
\displaystyle \therefore x\frac{dy}{dx} + a\frac{dx}{dy} = x\left(\frac{2a}{y}\right) + a\left(\frac{y}{2a}\right) \qquad \text{[Using equations (2) and (3)]}
\displaystyle \Rightarrow x\frac{dy}{dx} + a\frac{dx}{dy} = \frac{2ax}{y} + \frac{y}{2}
\displaystyle \Rightarrow x\frac{dy}{dx} + a\frac{dx}{dy} = \frac{y^{2}}{2y} + \frac{y}{2} \qquad \text{[Using equation (1)]}
\displaystyle \Rightarrow x\frac{dy}{dx} + a\frac{dx}{dy} = \frac{y}{2} + \frac{y}{2}
\displaystyle \Rightarrow x\frac{dy}{dx} + a\frac{dx}{dy} = y
\displaystyle \Rightarrow y = x\frac{dy}{dx} + a\frac{dx}{dy}
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 9: }~\text{Show that }Ax^{2}+By^{2}=1\text{ is a solution of the differential equation } \\ x\left\{y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}\right\}=y\frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle Ax^{2} + By^{2} = 1 \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle 2Ax + 2By\frac{dy}{dx} = 0 \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle 2A + 2B\left(\frac{dy}{dx}\right)^{2} + 2By\frac{d^{2}y}{dx^{2}} = 0
\displaystyle \Rightarrow 2B\left[y\frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2}\right] = -2A
\displaystyle \Rightarrow y\frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2} = -\frac{2A}{2B}
\displaystyle \Rightarrow y\frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2} = -\left(\frac{Ax}{By}\right)\frac{dy}{dx} \qquad \text{[Using equation (2)]}
\displaystyle \Rightarrow x\left[y\frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2}\right] = y\frac{dy}{dx}
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 10: }~\text{Show that }y=ax^{3}+bx^{2}+c\text{ is a solution of the differential equation } \\ \frac{d^{3}y}{dx^{3}}=6a.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = ax^{3} + bx^{2} + c \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = 3ax^{2} + 2bx \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = 6ax + 2b \qquad (3)
\displaystyle \text{Differentiating both sides of equation (3) with respect to } x, \text{ we get}
\displaystyle \frac{d^{3}y}{dx^{3}} = 6a
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 11: }~\text{Show that }y=\frac{c-x}{1+cx}\text{ is a solution of the differential equation } \\ (1+x^{2})\frac{dy}{dx}+(1+y^{2})=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = \frac{c - x}{1 + cx} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = \frac{(1+cx)(-1) - (c-x)c}{(1+cx)^{2}}
\displaystyle \Rightarrow \frac{dy}{dx} = \frac{-1 - cx - c^{2} + cx}{(1+cx)^{2}}
\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{1 + c^{2}}{(1+cx)^{2}} \qquad (2)
\displaystyle \text{Now,}
\displaystyle (1+x^{2})\frac{dy}{dx} + (1+y^{2})
\displaystyle = -(1+x^{2})\frac{1+c^{2}}{(1+cx)^{2}} + \left\{1 + \frac{(c-x)^{2}}{(1+cx)^{2}}\right\} \qquad \text{[Using (1) and (2)]}
\displaystyle = -\frac{(1+x^{2})(1+c^{2})}{(1+cx)^{2}} + \frac{(1+cx)^{2} + (c-x)^{2}}{(1+cx)^{2}}
\displaystyle = -\frac{(1+x^{2})(1+c^{2})}{(1+cx)^{2}} + \frac{1 + 2cx + c^{2}x^{2} + c^{2} - 2cx + x^{2}}{(1+cx)^{2}}
\displaystyle = -\frac{(1+x^{2})(1+c^{2})}{(1+cx)^{2}} + \frac{(1+x^{2}) + c^{2}(1+x^{2})}{(1+cx)^{2}}
\displaystyle = -\frac{(1+x^{2})(1+c^{2})}{(1+cx)^{2}} + \frac{(1+x^{2})(1+c^{2})}{(1+cx)^{2}}
\displaystyle = 0
\displaystyle \Rightarrow (1+x^{2})\frac{dy}{dx} + (1+y^{2}) = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 12: }~\text{Show that }y=e^{x}(A\cos x+B\sin x)\text{ is the solution of the differential equation } \\ \frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+2y=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = e^{x}(A\cos x + B\sin x) \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = e^{x}(A\cos x + B\sin x) + e^{x}(-A\sin x + B\cos x) \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = e^{x}(A\cos x + B\sin x) + e^{x}(-A\sin x + B\cos x) + e^{x}(-A\sin x + B\cos x) + e^{x}(-A\cos x - B\sin x)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 2e^{x}(-A\sin x + B\cos x)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 2e^{x}(-A\sin x + B\cos x) + 2e^{x}(A\cos x + B\sin x) - 2e^{x}(A\cos x + B\sin x)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = 2\frac{dy}{dx} - 2y \qquad \text{[Using equations (1) and (2)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 13: }~\text{Verify that }y=cx+2c^{2}\text{ is a solution of the differential equation } \\ 2\left(\frac{dy}{dx}\right)^{2}+x\frac{dy}{dx}-y=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = cx + 2c^{2} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = c \qquad (2)
\displaystyle \text{Now,}
\displaystyle 2\left(\frac{dy}{dx}\right)^{2} + x\frac{dy}{dx} - y
\displaystyle = 2c^{2} + cx - cx - 2c^{2} = 0 \qquad \text{[Using equations (1) and (2)]}
\displaystyle \Rightarrow 2\left(\frac{dy}{dx}\right)^{2} + x\frac{dy}{dx} - y = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 14: }~\text{Verify that }y=-x-1\text{ is a solution of the differential equation } \\ (y-x)\,dy-(y^{2}-x^{2})\,dx=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = -x - 1 \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = -1 \qquad (2)
\displaystyle \text{Now,}
\displaystyle \frac{dy}{dx} - \frac{y^{2} - x^{2}}{y - x}
\displaystyle = \frac{dy}{dx} - (y + x)
\displaystyle = -1 - (-x - 1 + x) \qquad \text{[Using equations (1) and (2)]}
\displaystyle = -1 + 1 = 0
\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y^{2} - x^{2}}{y - x}
\displaystyle \Rightarrow (y - x)\,dy = (y^{2} - x^{2})\,dx
\displaystyle \Rightarrow (y - x)\,dy - (y^{2} - x^{2})\,dx = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 15: }~\text{Verify that }y^{2}=4a(x+a)\text{ is a solution of the differential equations } \\ y\left\{1-\left(\frac{dy}{dx}\right)^{2}\right\}=2x\frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y^{2} = 4a(x+a) \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle 2y\frac{dy}{dx} = 4a
\displaystyle \Rightarrow y\frac{dy}{dx} = 2a
\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2a}{y} \qquad (2)
\displaystyle \text{Now,}
\displaystyle y\left\{1-\left(\frac{dy}{dx}\right)^{2}\right\} - 2x\frac{dy}{dx}
\displaystyle = y\left\{1-\frac{4a^{2}}{y^{2}}\right\} - 2x\left(\frac{2a}{y}\right)
\displaystyle = y\left(\frac{y^{2}-4a^{2}}{y^{2}}\right) - \frac{4ax}{y}
\displaystyle = \frac{y^{2}-4a^{2}}{y} - \frac{4ax}{y}
\displaystyle = \frac{(4ax+4a^{2})-4a^{2}}{y} - \frac{4ax}{y} \qquad \text{[Using equation (1)]}
\displaystyle = \frac{4ax}{y} - \frac{4ax}{y} = 0
\displaystyle \Rightarrow y\left\{1-\left(\frac{dy}{dx}\right)^{2}\right\} = 2x\frac{dy}{dx}
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 16: }~\text{Verify that }y=ce^{\tan^{-1}x}\text{ is a solution of the differential equation } \\ (1+x^{2})\frac{d^{2}y}{dx^{2}}+(2x-1)\frac{dy}{dx}=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = c e^{\tan^{-1}x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = c e^{\tan^{-1}x}\frac{1}{1+x^{2}} \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = c\frac{(1+x^{2})e^{\tan^{-1}x}\frac{1}{1+x^{2}} - e^{\tan^{-1}x}(2x)}{(1+x^{2})^{2}}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = c\frac{e^{\tan^{-1}x} - 2xe^{\tan^{-1}x}}{(1+x^{2})^{2}}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = c\frac{(1-2x)e^{\tan^{-1}x}}{(1+x^{2})^{2}}
\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}} = c(1-2x)\frac{e^{\tan^{-1}x}}{1+x^{2}}
\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}} = (1-2x)\frac{dy}{dx} \qquad \text{[Using equation (2)]}
\displaystyle \Rightarrow (1+x^{2})\frac{d^{2}y}{dx^{2}} + (2x-1)\frac{dy}{dx} = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 17: }~\text{Verify that }y=e^{m\cos^{-1}x}\text{ satisfies the differential equation } \\ (1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}-m^{2}y=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = e^{m\cos^{-1}x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = m e^{m\cos^{-1}x}\left(-\frac{1}{\sqrt{1-x^{2}}}\right)
\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{m e^{m\cos^{-1}x}}{\sqrt{1-x^{2}}} \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(-\frac{m e^{m\cos^{-1}x}}{\sqrt{1-x^{2}}}\right)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}} = (-m)\left[\frac{\sqrt{1-x^{2}}\, m e^{m\cos^{-1}x}\left(-\frac{1}{\sqrt{1-x^{2}}}\right) - e^{m\cos^{-1}x}\left(\frac{2x}{2\sqrt{1-x^{2}}}\right)}{1-x^{2}}\right]
\displaystyle \Rightarrow (1-x^{2})\frac{d^{2}y}{dx^{2}} = (-m)\left[-m e^{m\cos^{-1}x} + \frac{x e^{m\cos^{-1}x}}{\sqrt{1-x^{2}}}\right]
\displaystyle \Rightarrow (1-x^{2})\frac{d^{2}y}{dx^{2}} = m^{2} e^{m\cos^{-1}x} - m x \frac{e^{m\cos^{-1}x}}{\sqrt{1-x^{2}}}
\displaystyle \Rightarrow (1-x^{2})\frac{d^{2}y}{dx^{2}} = m^{2}y + x\frac{dy}{dx} \qquad \text{[Using equations (1) and (2)]}
\displaystyle \Rightarrow (1-x^{2})\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx} - m^{2}y = 0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 18: }~\text{Verify that }y=\log\left(x+\sqrt{x^{2}+a^{2}}\right)^{2}\text{ satisfies the differential equation } \\ (a^{2}+x^{2})\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=\log\left(x+\sqrt{x^{2}+a^{2}}\right)^{2} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[\log\left(x+\sqrt{x^{2}+a^{2}}\right)^{2}\right]
\displaystyle =\frac{d}{dx}\left[2\log\left(x+\sqrt{x^{2}+a^{2}}\right)\right]
\displaystyle =2\frac{1+\dfrac{x}{\sqrt{x^{2}+a^{2}}}}{x+\sqrt{x^{2}+a^{2}}}
\displaystyle =2\frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}}\left(x+\sqrt{x^{2}+a^{2}}\right)}
\displaystyle =\frac{2}{\sqrt{x^{2}+a^{2}}} \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=2\left(-\frac{1}{2}\right)\frac{2x}{(x^{2}+a^{2})\sqrt{x^{2}+a^{2}}}
\displaystyle \Rightarrow (a^{2}+x^{2})\frac{d^{2}y}{dx^{2}}=-\frac{2x}{\sqrt{x^{2}+a^{2}}}
\displaystyle \Rightarrow (a^{2}+x^{2})\frac{d^{2}y}{dx^{2}}=-x\frac{dy}{dx} \qquad \text{[Using equation (2)]}
\displaystyle \Rightarrow (a^{2}+x^{2})\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=0
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 19: }~\text{Show that the differential equation of which }y=2(x^{2}-1)+ce^{-x^{2}}\text{ is a solution is } \\ \frac{dy}{dx}+2xy=4x^{3}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = 2(x^{2}-1) + ce^{-x^{2}} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = 4x - c e^{-x^{2}}(2x)
\displaystyle = 2x\left[2 - c e^{-x^{2}}\right]
\displaystyle = -2x\left[2x^{2} - 2 + c e^{-x^{2}} - 2x^{2}\right]
\displaystyle = -2x\left[2(x^{2}-1) + c e^{-x^{2}} - 2x^{2}\right]
\displaystyle = -2x\left[y - 2x^{2}\right] \qquad \text{[Using equation (1)]}
\displaystyle \Rightarrow \frac{dy}{dx} = -2xy + 4x^{3}
\displaystyle \Rightarrow \frac{dy}{dx} + 2xy = 4x^{3}
\displaystyle \text{Hence, the given function is the solution of the given differential equation.}

\displaystyle \textbf{Question 20: }~\text{Show that }y=e^{-x}+ax+b\text{ is solution of the differential equation } \\ e^{x}\frac{d^{2}y}{dx^{2}}=1.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = e^{-x} + ax + b \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we have}
\displaystyle \frac{dy}{dx} = -e^{-x} + a \qquad (2)
\displaystyle \text{Differentiating both sides of equation (2) with respect to } x, \text{ we have}
\displaystyle \frac{d^{2}y}{dx^{2}} = e^{-x}
\displaystyle \Rightarrow e^{x}\frac{d^{2}y}{dx^{2}} = 1
\displaystyle \text{Hence, the given function is a solution of the given differential equation.}

\displaystyle \textbf{Question 21: }~\text{For each of the following differential equations verify that the} \\ \text{accompanying function is a solution:}

\displaystyle \text{(i): }x\frac{dy}{dx}=y;\ \hspace{5.0cm} y=ax.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y = ax \qquad (1)
\displaystyle \text{Given differential equation}
\displaystyle x\frac{dy}{dx} = y
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx} = a
\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y}{x} \qquad \text{[Using equation (1)]}
\displaystyle \Rightarrow x\frac{dy}{dx} = y
\displaystyle \text{Hence, the given function is the solution to the given differential equation.}

\displaystyle \text{(ii): }x+y\frac{dy}{dx}=0;\ \hspace{4.0cm}y=\pm\sqrt{a^{2}-x^{2}}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=\pm\sqrt{a^{2}-x^{2}}
\displaystyle \Rightarrow y^{2}=a^{2}-x^{2} \qquad (1)
\displaystyle \text{Given differential equation:}
\displaystyle x+y\frac{dy}{dx}=0
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle 2y\frac{dy}{dx}=-2x
\displaystyle \Rightarrow y\frac{dy}{dx}=-x
\displaystyle \Rightarrow x+y\frac{dy}{dx}=0
\displaystyle \text{Hence, the given function is the solution to the given differential equation.}

\displaystyle \text{(iii): }x\frac{dy}{dx}+y=y^{2};\ \hspace{4.0cm}y=\frac{a}{x+a}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=\frac{a}{x+a}
\displaystyle \Rightarrow xy+ay=a
\displaystyle \Rightarrow xy=a(1-y)
\displaystyle \Rightarrow \frac{xy}{1-y}=a
\displaystyle \Rightarrow \frac{1-y}{xy}=\frac{1}{a} \qquad (1)
\displaystyle \text{Given differential equation: } x\frac{dy}{dx}+y=y^{2}
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{xy(0-\frac{dy}{dx})-(1-y)\left(x\frac{dy}{dx}+y\right)}{(xy)^{2}}=0
\displaystyle \Rightarrow xy\left(-\frac{dy}{dx}\right)-(1-y)\left(x\frac{dy}{dx}+y\right)=0
\displaystyle \Rightarrow -xy\frac{dy}{dx}-x\frac{dy}{dx}-y+xy\frac{dy}{dx}+y^{2}=0
\displaystyle \Rightarrow -x\frac{dy}{dx}-y+y^{2}=0
\displaystyle \Rightarrow x\frac{dy}{dx}+y=y^{2}
\displaystyle \text{Hence, the given function is the solution to the given differential equation.}

\displaystyle \text{(iv): }x^{3}\frac{d^{2}y}{dx^{2}}=1;\ \hspace{4.0cm}y=ax+b+\frac{1}{2x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=ax+b+\frac{1}{2x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=a-\frac{1}{2x^{2}} \qquad (2)
\displaystyle \text{Now, differentiating both sides of equation (2) with respect to } x, \text{ we get}
\displaystyle \frac{d^{2}y}{dx^{2}}=\left(-\frac{1}{2}\right)\left(-\frac{2}{x^{3}}\right)
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=\frac{1}{x^{3}}
\displaystyle \Rightarrow x^{3}\frac{d^{2}y}{dx^{2}}=1
\displaystyle \text{Hence, the given function is the solution to the given differential equation.}

\displaystyle \text{(v): }y=\left(\frac{dy}{dx}\right)^{2};\ \hspace{4.0cm}y=\frac{1}{4}(x+a)^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=\frac{1}{4}(x+a)^{2} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{1}{4}\cdot 2(x+a)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{2}(x+a)
\displaystyle \text{Squaring both sides, we get}
\displaystyle \left(\frac{dy}{dx}\right)^{2}=\left[\frac{1}{2}(x+a)\right]^{2}
\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)^{2}=\frac{1}{4}(x+a)^{2}
\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)^{2}=y \qquad \text{[Using equation (1)]}
\displaystyle \therefore y=\left(\frac{dy}{dx}\right)^{2}
\displaystyle \text{Hence, the given function is the solution to the given differential equation.}

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