Initial & Boundary Value Problems

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Differential Equations › Initial & Boundary Value Problems

Questions 1 - 2

1

Solve the Boundary Value Problem (BVP).

$\y'+16y=0 \y(0)=-4 \\y\left(\frac{\pi}{2} \right )=5$

$y(x)=-4\cos(4x)+5\sin(4x)$

$y(x)=5\cos(4x)-4\sin(4x)$

$y(x)=4\cos(4x)-5\sin(4x)$

$y(x)=-5\cos(4x)+4\sin(4x)$

$y(x)=-4\cos(16x)+5\sin(16x)$

Explanation

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is,

$\y'+ay=0 \y(x)=c_1\cos(\sqrt{a}x)+c_2\sin(\sqrt{a}x)$

Therefore, the equation becomes

$y(x)=c_1\cos(4x)+c_2\sin(4x)$

From here, apply the boundary conditions to solve for the constants and

$\-4=y(0)=c_1 \\5=y\left(\frac{\pi}{2} \right )=c_2$

Thus resulting in the solution,

$y(x)=-4\cos(4x)+5\sin(4x)$

2

Solve the Boundary Value Problem (BVP).

$\y'+9y=0 \y(0)=-7 \\y\left(\frac{\pi}{2} \right )=2$

$y(x)=-7\cos(3x)+2\sin(3x)$

$y(x)=-2\cos(3x)+7\sin(3x)$

$y(x)=7\cos(3x)+2\sin(3x)$

$y(x)=-7\cos(3x)-2\sin(3x)$

$y(x)=-3\cos(7x)+3\sin(2x)$

Explanation

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is,

$\y'+ay=0 \y(x)=c_1\cos(\sqrt{a}x)+c_2\sin(\sqrt{a}x)$

Therefore, the equation becomes

$y(x)=c_1\cos(3x)+c_2\sin(3x)$

From here, apply the boundary conditions to solve for the constants and

$\-7=y(0)=c_1 \\2=y\left(\frac{\pi}{2} \right )=c_2$

Thus resulting in the solution,

$y(x)=-7\cos(3x)+2\sin(3x)$

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