Consider the IVP{ y' = 2y{y(0) = 1a) Find the exact solution of the IVPb) Using step ℎ = 0.2, find an approximate solution of the PVI at point = 1.0b1) Euler's methodb2) Improved Euler Methodb3) Fourth-order Runge-Kutta methodc) Compare the approximate solutions obtained in the previous items with the exact solution.

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Consider the IVP{ y&#039; = 2y{y(0) = 1a) Find the exact solution of the IVPb) Using step ℎ = 0.2, find an approximate solution of the PVI at point  = 1.0b1) Euler&#039;s methodb2) Improved Euler Methodb3) Fourth-order Runge-Kutta methodc) Compare the approximate solutions obtained in the previous items with the exact solution.

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To find the exact solution of the IVP&amp;nbsp;{y&#039;=2y,{y(0)=1, we can start by separating the variables and then integrating both sides.dydx=2ydyy=2dxIntegrating both sides, we get:ln|y|=2x+C1&amp;nbsp; &amp;nbsp; (where&amp;nbsp;C1&amp;nbsp;is the constant of integration)To solve for y, we exponentiate both sides:|y|=e2x+C1=e2x⋅eC1Since&amp;nbsp;y(0)=1, we can substitute these values into our equation to solve for the constant of integration&amp;nbsp;C1:|1|=e2⋅0+C1=eC1Taking the natural logarithm of both sides:ln|1|=ln|eC1|0=C1Substituting this value back into our previous equation, we get:|y|=e2xSince&amp;nbsp;y(0)=1, we know that y must be positive. Therefore, we can drop the absolute value bars and write the solution as:y=e2xTherefore, the exact solution of the IVP&amp;nbsp;{y&#039;=2y,{y(0)=1&amp;nbsp;is&amp;nbsp;y=e2x.

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Consider the IVP { y' = 2y {y(0) = 1 a)

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