Peurre ypabrente Broporo nopstitka: y''=2x^2-3x+1 y(0)=0,y'(0)=4

To solve the given second-order differential equation with initial conditions, we need to integrate the equation twice and apply the initial conditions.<br /><br />Given:<br />$y'' = 2x^2 - 3x + 1$<br />$y(0) = 0$<br />$y'(0) = 4$<br /><br />Step 1: Integrate $y''$ to find $y'$.<br />$\int (2x^2 - 3x + 1) dx = \int y'' dx$<br />$y' = \int (2x^2 - 3x + 1) dx$<br />$y' = \frac{2}{3}x^3 - \frac{3}{2}x^2 + x + C_1$<br /><br />Step 2: Apply the initial condition $y'(0) = 4$ to find $C_1$.<br />$y'(0) = 4$<br />$\frac{2}{3}(0)^3 - \frac{3}{2}(0)^2 + 0 + C_1 = 4$<br />$C_1 = 4$<br /><br />So, $y' = \frac{2}{3}x^3 - \frac{3}{2}x^2 + x + 4$<br /><br />Step 3: Integrate $y'$ to find $y$.<br />$\int (\frac{2}{3}x^3 - \frac{3}{2}x^2 + x + 4) dx = \int y' dx$<br />$y = \frac{1}{2}x^4 - \frac{3}{4}x^3 + \frac{1}{2}x^2 + 4x + C_2$<br /><br />Step 4: Apply the initial condition $y(0) = 0$ to find $C_2$.<br />$y(0) = 0$<br />$\frac{1}{2}(0)^4 - \frac{3}{4}(0)^3 + \frac{1}{2}(0)^2 + 4(0) + C_2 = 0$<br />$C_2 = 0$<br /><br />So, $y = \frac{1}{2}x^4 - \frac{3}{4}x^3 + \frac{1}{2}x^2 + 4x$<br /><br />Therefore, the solution to the given second-order differential equation with initial conditions is:<br />$y = \frac{1}{2}x^4 - \frac{3}{4}x^3 + \frac{1}{2}x^2 + 4x$