Initial & Boundary Value Problems - Partial Differential Equations

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Question

Solve the Boundary Value Problem (BVP).

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

Answer

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,

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

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Question

Solve the Boundary Value Problem (BVP).

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

Answer

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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