Question
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0,\text{y}(0)=1,\text{y}'(0)=1$Function $\text{y}=\sin\text{x}+\cos\text{x}$
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Answer
We have,
$\text{y}=\sin\text{x}+\cos\text{x}\ ....(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\cos\text{x}-\sin{\text{x}}...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\sin\text{x}-\cos\text{x}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-(\sin\text{x}+\cos\text{x})$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y}$ [Using (1)]
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
It is the given differential equation.
Therefore, $\text{y}=\sin\text{x}+\cos\text{x}$ satiesfies the given differential equation.
Also, when $\text{x}=0;\text{y}=\sin0+\cos0=1,\text{i.e.y}(0)=1.$
And, when $\text{x}=0;\text{y}'=\cos0-\sin0=1,\text{i.e.y'}(0)=1$
Hence $\text{y}=\sin\text{x}+\cos\text{x}$ is the solution to the given initial value problem.
$\text{y}=\sin\text{x}+\cos\text{x}\ ....(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\cos\text{x}-\sin{\text{x}}...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\sin\text{x}-\cos\text{x}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-(\sin\text{x}+\cos\text{x})$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y}$ [Using (1)]
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
It is the given differential equation.
Therefore, $\text{y}=\sin\text{x}+\cos\text{x}$ satiesfies the given differential equation.
Also, when $\text{x}=0;\text{y}=\sin0+\cos0=1,\text{i.e.y}(0)=1.$
And, when $\text{x}=0;\text{y}'=\cos0-\sin0=1,\text{i.e.y'}(0)=1$
Hence $\text{y}=\sin\text{x}+\cos\text{x}$ is the solution to the given initial value problem.
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