Question

The solution for the differential equation \(\dfrac{d^2y}{dx^2}=3x-2\) with boundary conditions \(y(0)=2\) and \(y'(1)=-3\) is

• A. \(y=\dfrac{x^3}{3}-\dfrac{x^2}{2}-3x-2\)
• B. \(y=3x^3-\dfrac{x^2}{2}+5x+\dfrac{3}{2}\)
• C. \(y=3x^3-\dfrac{x^2}{2}-5x+2\)
• D. \(y=\dfrac{x^3}{2}-x^2-\dfrac{5x}{2}+2\)

Hint:

When \(y''\) is given, integrate twice and use the given conditions to find the two constants.

Answer 1

The Correct Option is D

Concept:
If \[ \frac{d^2y}{dx^2}=f(x) \] then we integrate once to get \[ \frac{dy}{dx} \] and integrate again to get \[ y \] The constants of integration are found using the given boundary conditions.

Step 1: Write the given differential equation. \[ \frac{d^2y}{dx^2}=3x-2 \]

Step 2: Integrate once. \[ \frac{dy}{dx}=\int(3x-2)\,dx \] \[ \frac{dy}{dx}=\frac{3x^2}{2}-2x+C_1 \]

Step 3: Use the condition \(y'(1)=-3\). Since \[ y'=\frac{dy}{dx} \] we substitute \(x=1\): \[ -3=\frac{3(1)^2}{2}-2(1)+C_1 \] \[ -3=\frac{3}{2}-2+C_1 \] \[ -3=-\frac{1}{2}+C_1 \] \[ C_1=-3+\frac{1}{2} \] \[ C_1=-\frac{5}{2} \]

Step 4: Substitute \(C_1\) in \(y'\). \[ \frac{dy}{dx}=\frac{3x^2}{2}-2x-\frac{5}{2} \]

Step 5: Integrate again to find \(y\). \[ y=\int\left(\frac{3x^2}{2}-2x-\frac{5}{2}\right)\,dx \] \[ y=\frac{x^3}{2}-x^2-\frac{5x}{2}+C_2 \]

Step 6: Use the condition \(y(0)=2\). Substitute \(x=0\): \[ 2=\frac{0^3}{2}-0^2-\frac{5(0)}{2}+C_2 \] \[ 2=C_2 \]

Step 7: Write the final solution. \[ y=\frac{x^3}{2}-x^2-\frac{5x}{2}+2 \]

Step 8: Final answer. \[ \boxed{y=\frac{x^3}{2}-x^2-\frac{5x}{2}+2} \]