Question

The solution for the following differential equation with boundary conditions $y\left(0\right)=2$ and $y^{\prime }\left(1\right)=-3$ is, $\frac{d^{2}y}{d{x}^2}=3x-2$

• A. $y=\frac{x^3}{3}-\frac{x^2}{2}+3x-6$
• B. $y=3x^3-\frac{x^2}{2}-5x+2$
• C. $y=\frac{x^3}{2}-x^2-\frac{5x}{2}+2$
• D. $y=x^3-\frac{x^2}{2}+5x+\frac{3}{2}$

Answer 1

The correct option is C $y=\frac{x^3}{2}-x^2-\frac{5x}{2}+2$

$\frac{d^{2}y}{d{x}^2}=3x-2$
Interating both sides w.r.t. x
$\Rightarrow$ $\frac{dy}{dx}=\frac{3}{2}{x}^2-2x+C_1$
At $x=1,$ $\frac{dy}{dx}=-3$
$\Rightarrow$ $-3=\frac{3}{2}-2+C_1$
$\Rightarrow$ $C_1=-\frac{5}{2}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{3}{2}{x}^2-2x-\frac{5}{2}$
Again, integrating both sides w.r.t. x
$\Rightarrow$ $y=\frac{x^3}{2}-x^2-\frac{5}{2}x+C_2$
At $x=0,y=2$
$\Rightarrow$ $C_2=2$
$\Rightarrow$ $y=\frac{x^3}{2}-x^2-\frac{5}{2}x+2$