Question

solve: (x^2d^2+xd-1)y=x^3/1+x^2​

Answer 1

Answer:

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Answer 2

The solution to the given differential equation is $y = C_1 x + \dfrac{C_2}{x} + \Big( \dfrac{x^{2} -1}{4x} \Big) log(x^{2} + 1) + \Big( \dfrac{x^{2} + 1}{4x} \Big)$

Step-by-step explanation:

Given: The differential equation $(x^2D^2+xD-1)y=\frac{x^3}{1+x^2}$

To Find: The solution to the given differential equation

Solution:

• Finding the solution of the given differential equation

The given differential equation can be solved by using the method of variation of parameters.

For the given differential equation, let $x = e^z$ such that we have $x \frac{dy}{dx} = Dy$ and $x^{2} \frac{d^{2} y}{dx^{2} } = D(D-1)y$ where $D = \frac{d}{dz}$

Therefore, the differential equation becomes,

$\Rightarrow (D^{2} - 1)y = \dfrac{e^{3z}}{1+e^{2z}}$

The complementary function ( $y_{_{CF}} = C_1 y_1 + C_2 y_2$ ) is given by,

$\Rightarrow m^2 - 1 = 0 \Rightarrow m = \pm 1$

$\Rightarrow y_{_{CF}} = C_1 e^z + C_2 e^{-z}$

The particular integral is given as $y_{_{PI}} = u y_1 + v y_2$ where $y_1 \ \text{and } y_2$ can be found from the above expression, i.e. $y_1 = e^z$ and $y_2 = e^{-z}$; $u \text{ and } v$ can be determined by the below formulas.

$u = - \int \dfrac{y_2 f(z)}{y_1 y'_2 - y'_1 y_2} dz \text{ and } v = \int \dfrac{y_1 f(z)}{y_1 y'_2 - y'_1 y_2} dz$

Therefore, we have,

$\Rightarrow u = - \int \dfrac{(e^{-z}) \Big( \dfrac{e^{3z}}{1+e^{2z}} \Big)}{(e^{z}) (-e^{-z}) - (e^{z}) (e^{-z})} dz$

$\Rightarrow u = - \int \dfrac{e^{2z}}{( {1+e^{2z}}) (-2)} dz$

To solve the above integral, let $t = 1 + e^{2z} \text{ such that } \Rightarrow dt = 2 e^{2z} dz$, therefore,

$\Rightarrow u = \dfrac{1}{4} \int \dfrac{1}{t} dt = \dfrac{1}{4} log (t)$

$\Rightarrow u = \dfrac{1}{4} log(1 + e^{2z})$

Similarly, $v = \dfrac{1}{4} [ (1 + e^{2z}) - log (1 + e^{2z})]$

Therefore, the particular integral is $y_{_{PI}} = u y_1 + v y_2 = \dfrac{e^{z}}{4} [ log (1 + e^{2z})] + \dfrac{e^{-z}}{4} [ (1 + e^{2z}) - log (1 + e^{2z})]$

The complete solution of the differential equation is $y = y_{_{CF}} + y_{_{PI}}$

$\Rightarrow y = C_1 e^{z} + {C_2 e^{-z}} + \Big( \dfrac{e^{2z} -1}{4e^{z}} \Big) log(e^{2z} + 1) + \Big( \dfrac{e^{2z} + 1}{4e^{z}} \Big)$

Now, putting $e^{z} = x$ in the above expression, we get

$\Rightarrow y = C_1 x + \dfrac{C_2}{x} + \Big( \dfrac{x^{2} -1}{4x} \Big) log(x^{2} + 1) + \Big( \dfrac{x^{2} + 1}{4x} \Big)$

Hence, the solution to the given differential equation is $y = C_1 x + \dfrac{C_2}{x} + \Big( \dfrac{x^{2} -1}{4x} \Big) log(x^{2} + 1) + \Big( \dfrac{x^{2} + 1}{4x} \Big)$