The solution of Tx" + (4t - 2)X' + (13t - 4)X = 0, If X(0) = 0 is mathematically given as
[tex]&n \times(S)=\frac{c}{\left((s+2)^{2}+9\right)^{2}}[/tex]
Generally, the equation for is mathematically given as
[tex]&t x^{\prime \prime}+(4 t-2) x^{\prime}+(13 t-4) x=0, \quad x(0)=0\\\\&\Rightarrow \quad t x^{\prime \prime}+4 t x^{\prime}-2 x^{i}+13 t x-4 x=5\\[/tex]
By taking Laplace to transform,
[tex]&L\left\{t x^{13}\right\}+L\left\{4 t x^{\prime}\right\}-L\left\{2 x^{\prime}\right\}+L\{13 t x\}-L\{4 x\}=0\\\\\\&-25 x(s)-s^{2} x^{1}(5)-4 x(s)-45 x^{\prime}(s)-25 x(s)-13 x^{\prime}(5)-4 x(s)=0\\[/tex]
[tex]&\Rightarrow \quad-\left(5^{2}+45+13\right) x^{4}(5)-4(5+2) x(5)=0\\\\\\&\Rightarrow \quad\left(s^{2}+45+13\right) x^{1}(5)+4(5+2) x(5)=0\\\\\\&\Rightarrow\left(s^{2}+4 s+13\right) x^{\prime}(s)=-4(s+2)^{x}(s)\\\\\\&\Rightarrow \frac{x^{\prime}(s)}{x(s)}=-\frac{4(s+2)}{s^{2}+4 s+13}\\\\\[/tex]
In conclusion, By integrating both sides
The solution is
[tex]&x(s)=\frac{c}{\left(s^{2}+4 s+13\right)^{2}}\\[/tex]
[tex]&n \times(S)=\frac{c}{\left((s+2)^{2}+9\right)^{2}}[/tex]
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