3araya 2. Penure ypastiente Broporo nopsuka: y''=2x^2-3x+1 y(0)=0,y'(0)=4

To solve the given differential equation with the given initial conditions, we can use the method of integrating factors.<br /><br />The given differential equation is:<br />$y'' = 2x^2 - 3x + 1$<br /><br />We can rewrite this as:<br />$y'' - 2x^2 + 3x - 1 = 0$<br /><br />Now, we can find the integrating factor, which is given by:<br />$\mu(x) = e^{\int P(x) dx}$<br /><br />where $P(x)$ is the coefficient of $x$ in the linear term of the differential equation.<br /><br />In this case, $P(x) = 3x - 2$, so:<br />$\mu(x) = e^{\int (3x - 2) dx} = e^{\frac{3x^2}{2} - 2x}$<br /><br />Now, we can multiply the original differential equation by the integrating factor:<br />$\mu(x) y'' - \mu(x) (2x^2 - 3x + 1) = 0$<br /><br />Simplifying, we get:<br />$\mu(x) y'' - 2\mu(x) x^2 + 3\mu(x) x - \mu(x) = 0$<br /><br />Now, we can integrate both sides with respect to $x$:<br />$\int \mu(x) y'' dx = \int (2\mu(x) x^2 - 3\mu(x) x + \mu(x)) dx$<br /><br />Simplifying, we get:<br />$\mu(x) y' = \int (2\mu(x) x^2 - 3\mu(x) x + \mu(x)) dx$<br /><br />Now, we can use the initial conditions to find the value of the constant of integration:<br />$y(0) = 0$ and $y'(0) = 4$<br /><br />Substituting these values, we get:<br />$\mu(0) y'(0) = \int (2\mu(0) x^2 - 3\mu(0) x + \mu(0)) dx$<br /><br />Simplifying, we get:<br />$4\mu(0) = \int (2\mu(0) x^2 - 3\mu(0) x + \mu(0)) dx$<br /><br />Now, we can integrate both sides with respect to $x$:<br />$4\mu(0) = \frac{2\mu(0) x^3}{3} - \frac{3\mu(0) x^2}{2} + \mu(0) x + C$<br /><br />Simplifying, we get:<br />$4\mu(0) = \frac{2\mu(0) x^3}{3} - \frac{3\mu(0) x^2}{2} + \mu(0) x + C$<br /><br />Now, we can solve for the constant of integration $C$:<br />$C = 4\mu(0) - \frac{2\mu(0) x^3}{3} + \frac{3\mu(0) x^2}{2} - \mu(0) x$<br /><br />Finally, we can substitute the value of $C$ back into the equation for $y'$ and solve for $y$:<br />$y' = \int (2\mu(x) x^2 - 3\mu(x) x + \mu(x)) dx$<br /><br />Simplifying, we get:<br />$y' = \frac{2\mu(x) x^3}{3} - \frac{3\mu(x) x^2}{2} + \mu(x) x + C$<br /><br />Now, we can integrate both sides with respect to $x$ to find the general solution for $y$:<br />$y = \int (\frac{2\mu(x) x^3}{3} - \frac{3\mu(x) x^2}{2} + \mu(x) x + C) dx$<br /><br />Simplifying, we get:<br />$y = \frac{2\mu(x) x^4}{12} - \frac{3\mu(x) x^3}{6} + \frac{\mu(x) x^2}{2} + Cx + D$<br /><br />Now, we can use the initial conditions to find the values of the constants $C$ and $D$:<br />$y(0) = 0$ and $y'(0) = 4$<br /><br />Substituting these values, we get:<br />$0 = \frac{2\mu(0) 0^4}{12} - \frac{3\mu(0) 0^3}{6} + \frac{\mu(0) 0^2}{2} + C \cdot 0 + D$<br /><br />Simplifying, we get:<br />$0 = D$<br /><br />Now, we can substitute the value of $D$