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

\displaystyle \textbf{Question 1: }~x\frac{dy}{dx}=1,\ y(1)=0; \hspace{5.0cm} y=\log x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=\log x \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\frac{1}{x}
\displaystyle \text{or,}
\displaystyle x\frac{dy}{dx}=1
\displaystyle \text{It is the given differential equation.}
\displaystyle \text{Thus, } y=\log x \text{ satisfies the given differential equation.}
\displaystyle \text{Hence, it is a solution.}
\displaystyle \text{Also, when } x=1,\ y=\log 1=0,\ \text{i.e., } y(1)=0
\displaystyle \text{Hence, } y=\log x \text{ is the solution to the given initial value problem.}

\displaystyle \textbf{Question 2: }~\frac{dy}{dx}=y,\ y(0)=1; \hspace{5.0cm} y=e^{x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=e^{x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=e^{x}
\displaystyle \Rightarrow \frac{dy}{dx}=y \qquad \text{[Using equation (1)]}
\displaystyle \text{It is the given differential equation.}
\displaystyle \text{Here, } y=e^{x} \text{ satisfies the given differential equation, hence it is a solution.}
\displaystyle \text{Also, when } x=0,\ y=e^{0}=1,\ \text{i.e., } y(0)=1
\displaystyle \text{Hence, } y=e^{x} \text{ is the solution to the given initial value problem.}

\displaystyle \textbf{Question 3: }~\frac{d^{2}y}{dx^{2}}+y=0,\ y(0)=0,\ y'(0)=1; \hspace{2.0cm} y=\sin x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=\sin x \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\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}}=-\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{It is the given differential equation.}
\displaystyle \text{Here, } y=\sin x \text{ satisfies the given differential equation; hence, it is a solution.}
\displaystyle \text{Also, when } x=0,\ y=\sin 0=0,\ \text{i.e., } y(0)=0
\displaystyle \text{And, when } x=0,\ y'=\cos 0=1,\ \text{i.e., } y'(0)=1
\displaystyle \text{Hence, } y=\sin x \text{ is the solution to the given initial value problem.}

\displaystyle \textbf{Question 4: }~\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}=0,\ y(0)=2,\ y'(0)=1; \hspace{2.0cm} y=e^{x}+1.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=e^{x}+1 \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=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}}=e^{x}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=\frac{dy}{dx} \qquad \text{[Using equation (2)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}=0
\displaystyle \text{It is the given differential equation.}
\displaystyle \text{Here, } y=e^{x}+1 \text{ satisfies the given differential equation; hence, it is a solution.}
\displaystyle \text{Also, when } x=0,\ y=e^{0}+1=1+1=2,\ \text{i.e., } y(0)=2
\displaystyle \text{And, when } x=0,\ y'=e^{0}=1,\ \text{i.e., } y'(0)=1
\displaystyle \text{Hence, } y=e^{x}+1 \text{ is the solution to the given initial value problem.}

\displaystyle \textbf{Question 5: }~\frac{dy}{dx}+y=2,\ y(0)=3; \hspace{4.0cm} y=e^{-x}+2.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=e^{-x}+2 \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=-e^{-x}
\displaystyle \Rightarrow \frac{dy}{dx}=-(y-2) \qquad \text{[Using equation (1)]}
\displaystyle \Rightarrow \frac{dy}{dx}+y=2
\displaystyle \text{It is the given differential equation.}
\displaystyle \text{Here, } y=e^{-x}+2 \text{ satisfies the given differential equation; hence, it is a solution.}
\displaystyle \text{Also, when } x=0,\ y=e^{0}+2=1+2=3,\ \text{i.e., } y(0)=3
\displaystyle \text{Hence, } y=e^{-x}+2 \text{ is the solution to the given initial value problem.}

\displaystyle \textbf{Question 6: }~\frac{d^{2}y}{dx^{2}}+y=0,\ y(0)=1,\ y'(0)=1; \hspace{2.0cm} y=\sin x+\cos x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=\sin x+\cos x \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=\cos x-\sin 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}}=-\sin x-\cos x
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=-(\sin x+\cos 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{It is the given differential equation.}
\displaystyle \text{Therefore, } y=\sin x+\cos x \text{ satisfies the given differential equation.}
\displaystyle \text{Also, when } x=0,\ y=\sin 0+\cos 0=1,\ \text{i.e., } y(0)=1
\displaystyle \text{And, when } x=0,\ y'=\cos 0-\sin 0=1,\ \text{i.e., } y'(0)=1
\displaystyle \text{Hence, } y=\sin x+\cos x \text{ is the solution to the given initial value problem.}

\displaystyle \textbf{Question 7: }~\frac{d^{2}y}{dx^{2}}-y=0,\ y(0)=2,\ y'(0)=0; \hspace{2.0cm} y=e^{x}+e^{-x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=e^{x}+e^{-x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=e^{x}-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}}=e^{x}+e^{-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{It is the given differential equation.}
\displaystyle \text{Therefore, } y=e^{x}+e^{-x} \text{ satisfies the given differential equation.}
\displaystyle \text{Also, when } x=0,\ y=e^{0}+e^{0}=1+1=2,\ \text{i.e., } y(0)=2
\displaystyle \text{And, when } x=0,\ y'=e^{0}-e^{0}=1-1=0,\ \text{i.e., } y'(0)=0
\displaystyle \text{Hence, } y=e^{x}+e^{-x} \text{ is the solution to the given initial value problem.}

\displaystyle \textbf{Question 8: }~\frac{d^{2}y}{dx^{2}}-3\frac{dy}{dx}+2y=0,\ y(0)=2,\ y'(0)=3; \hspace{1.0cm} y=e^{x}+e^{2x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=e^{x}+e^{2x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=e^{x}+2e^{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}}=e^{x}+4e^{2x}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=3(e^{x}+2e^{2x})-2(e^{x}+e^{2x})
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=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{It is the given differential equation.}
\displaystyle \text{Therefore, } y=e^{x}+e^{2x} \text{ satisfies the given differential equation.}
\displaystyle \text{Also, when } x=0,\ y=e^{0}+e^{0}=1+1=2,\ \text{i.e., } y(0)=2
\displaystyle \text{And, when } x=0,\ y'=e^{0}+2e^{0}=1+2=3,\ \text{i.e., } y'(0)=3
\displaystyle \text{Hence, } y=e^{x}+e^{2x} \text{ is the solution to the given initial value problem.}

\displaystyle \textbf{Question 9: }~\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+y=0,\ y(0)=1,\ y'(0)=2; \hspace{1.0cm} y=xe^{x}+e^{x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y=xe^{x}+e^{x} \qquad (1)
\displaystyle \text{Differentiating both sides of equation (1) with respect to } x, \text{ we get}
\displaystyle \frac{dy}{dx}=xe^{x}+e^{x}+e^{x}
\displaystyle \Rightarrow \frac{dy}{dx}=xe^{x}+2e^{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}}=xe^{x}+e^{x}+2e^{x}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=xe^{x}+3e^{x}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=2(xe^{x}+2e^{x})-(xe^{x}+e^{x})
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}=2\frac{dy}{dx}-y \qquad \text{[Using equations (1) and (2)]}
\displaystyle \Rightarrow \frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+y=0
\displaystyle \text{It is the given differential equation.}
\displaystyle \text{Thus, } y=xe^{x}+e^{x} \text{ satisfies the given differential equation.}
\displaystyle \text{Also, when } x=0,\ y=0+1=1,\ \text{i.e., } y(0)=1
\displaystyle \text{And, when } x=0,\ y'=0+2=2,\ \text{i.e., } y'(0)=2
\displaystyle \text{Hence, } y=xe^{x}+e^{x} \text{ is the solution to the given initial value problem.}

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