Differential equation d^2y dx^2 -2 dy dx +y=0,y(0)=1,y(0)=2 Function y=xe^x+e^ x

We have, y=xe^x+e^ x ...(1) Differentiating both sides of (1) with respect to x, we get dy dx =xe^ x +e^ x +e^ x dy dx =xe^ x +2e^ x ...(2) Differentiating both sides of (2) with respect to x, we get d^ 2 y dx^ 2 =xe^ x +e^ x +2e^ x d^ 2 y dx^ 2 =xe^ x +3e^ x d^ 2 y dx^ 2 =2(xe^ x +2e^ x )(xe^ x +e^ x ) d^ 2 y dx^ 2 =2 dy dx -y [Using(1)and(2)] d^ 2 y dx^ 2 -2 dy dx +y=0 d^ 2 y dx^ 2 -2 dy dx +y=0 It is the given differential equation. Thus, y = xe^x + e^x satisfies the given differential equation. Also, when x = 0, y = 0 + 1 = 1, i.e. y(0) = 1 And, when x = 0, y' = 0 + 2 = 2, i.e. y'(0) = 2 Hence, y = xe^x + e^x is the solution to the given initialvalue problem.

Differential equation d^2y dx^2 -2 dy dx +y=0,y(0)=1,y(0)=2 Funct…

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Question

Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=1,\text{y}(0)=2$
Function $\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}$

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Answer

We have,
$\text{y}=\text{xe}^\text{x}+\text{e}^{\text{x}}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{xe}^{\text{x}}+\text{e}^{\text{x}}+\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{xe}^{\text{x}}+2\text{e}^{\text{x}}...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{xe}^{\text{x}}+\text{e}^{\text{x}}+2\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{xe}^{\text{x}}+3\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=2(\text{xe}^{\text{x}}+2\text{e}^{\text{x}})(\text{xe}^{\text{x}}+\text{e}^{\text{x}})$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=2\frac{\text{dy}}{\text{dx}}-\text{y}$ [Using(1)and(2)]
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-2\frac{\text{dy}}{\text{dx}}+\text{y}=0$
It is the given differential equation.
Thus, $y = xe^x + e^x$
satisfies the given differential equation.
Also, when $x = 0, y = 0 + 1 = 1,$ i.e. $y(0) = 1$
And, when $x = 0, y' = 0 + 2 = 2,$ i.e. $y'(0) = 2$
Hence, $y = xe^x + e^x$ is the solution to the given initialvalue problem.

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