# Using Laplave Transform, evaluate the integro-differential equation. y"(x)+9y(x)=40e^x . y(0)=5 , y'(0)=-2

Using Laplave Transform, evaluate the integro-differential equation
$$y''(x)+9y(x)=40e^x ; y(0)=5 , y'(0)=-2$$

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izboknil3

Step 1
Solve the given differential equation by Laplace transform.
The given equation is $$y''(x)+9y(x)=40e^x ; y(0)=5 , y'(0)=-2$$
The given equation is $$\displaystyle{y}{\left({0}\right)}={5}\ \text{ and }\ {y}'{\left({0}\right)}=-{2}$$
Step 2
Taking the Laplace transform on both sides of given equation.
$$L\left\{y''(x)+9y(x)\right\}=L\left\{40e^x\right\}$$
$$L\left\{y''(x)\right\}+9L\left\{y(x)\right\}=40L(e^x)$$
$$\left\{s^2\overline{y}-sy(0)-y'(0)\right\}+9\overline{y}=40\left\{\frac{1}{s-1}\right\} \ \ \left\{L(e^{ax})=\frac{1}{s-a}\right\}$$
$$s^2\overline{y}-s(5)-(-2)+9\overline{y}=\frac{40}{s-1} \ \ \ \left\{y(0)=5 \text{ and } y'(0)=-2\right\}$$
$$\displaystyle{s}^{{2}}\overline{{{y}}}-{5}{s}+{2}+{9}\overline{{{y}}}={\frac{{{40}}}{{{s}-{1}}}}$$
$$\displaystyle\overline{{{y}}}{\left({s}^{{2}}+{9}\right)}={\frac{{{40}}}{{{s}-{1}}}}+{5}{s}-{2}$$
$$\displaystyle\overline{{{y}}}{\left({s}^{{2}}+{9}\right)}={\frac{{{40}+{\left({5}{s}-{2}\right)}{\left({s}-{1}\right)}}}{{{s}-{1}}}}$$
$$\displaystyle\overline{{{y}}}={\frac{{{5}{s}^{{2}}-{7}{s}+{42}}}{{{\left({s}-{1}\right)}{\left({s}^{{2}}+{9}\right)}}}}\dot{{s}}{\left({2}\right)}$$
Step 3
Solve the R.H.S. of equation (2) by partial fraction.
$$\displaystyle{\frac{{{5}{s}^{{2}}-{7}{s}+{42}}}{{{\left({s}-{1}\right)}{\left({s}^{{2}}+{9}\right)}}}}={\frac{{{A}}}{{{s}-{1}}}}+{\frac{{{B}{s}+{C}}}{{{s}^{{2}}+{9}}}}$$
$$\displaystyle={\frac{{{A}{\left({s}^{{2}}+{9}\right)}+{\left({B}{s}+{C}\right)}{\left({s}-{1}\right)}}}{{{\left({s}-{1}\right)}{\left({s}^{{2}}+{9}\right)}}}}$$
$$\displaystyle={\frac{{{A}{s}^{{2}}+{9}{A}+{B}{s}^{{2}}-{B}{s}+{C}{s}-{C}}}{{{\left({s}-{1}\right)}{\left({s}^{{2}}+{9}\right)}}}}$$
$$\displaystyle={\frac{{{\left({A}+{B}\right)}{s}^{{2}}+{\left(-{B}+{C}\right)}{s}+{\left({9}{A}-{C}\right)}}}{{{\left({s}-{1}\right)}{\left({s}^{{2}}+{9}\right)}}}}$$
Equating the coefficients of same variables.
$$A+B=5$$
$$-B+C=-7$$
$$9A-C=42$$
Hence, $$a=5 , c=-6 , b=1$$
$$\displaystyle{\frac{{{5}{s}^{{2}}-{7}{s}+{42}}}{{{\left({s}-{1}\right)}{\left({s}^{{2}}+{9}\right)}}}}={\frac{{{4}}}{{{s}-{1}}}}+{\frac{{{s}-{6}}}{{{s}^{{2}}+{9}}}}\dot{{s}}{\left({3}\right)}$$
Step 4
Equating equation (2) and (3).
$$\overline{y}=\frac{4}{s-1}+\frac{s-6}{s^2+9} \left\{\text{ from (2) and (3)}\right\}$$
Taking the inverse transform
$$L^{-1}\left\{\overline{y}\right\}=L^{-1}\left\{\frac{4}{s-1}+\frac{s-6}{s^2+9}\right\}$$
$$y=4L^{-1}\left\{\frac{1}{s-1}\right\}+L^{-1}\left\{\frac{s}{s^2+3^2}\right\}-6L^{-1}\left\{\frac{1}{s^2+3^2}\right\}$$
$$\displaystyle={4}{e}^{{x}}+{\cos{{3}}}{x}-{6}\cdot{\frac{{{1}}}{{{3}}}}{\sin{{3}}}{x}$$
$$\displaystyle={4}{e}^{{x}}+{\cos{{3}}}{x}-{2}{\sin{{3}}}{x}$$
Hence, the solution of the equation is $$\displaystyle{4}{e}^{{x}}+{\cos{{3}}}{x}-{2}{\sin{{3}}}{x}$$