Initial & Boundary Value Problems Flashcards - Differential Equations

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,

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

Compare your answer with the correct one above

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,

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

Compare your answer with the correct one above

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,

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

Compare your answer with the correct one above

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,

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

Compare your answer with the correct one above

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,

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

Compare your answer with the correct one above

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,

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

Compare your answer with the correct one above

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,

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

Compare your answer with the correct one above

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,

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

Compare your answer with the correct one above

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Initial & Boundary Value Problems - Differential Equations

Solve the Boundary Value Problem (BVP).

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is, Therefore, the equation becomes From here, apply the boundary conditions to solve for the constants and Thus resulting in the solution,

Solve the Boundary Value Problem (BVP).

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is, Therefore, the equation becomes From here, apply the boundary conditions to solve for the constants and Thus resulting in the solution,

Solve the Boundary Value Problem (BVP).

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is, Therefore, the equation becomes From here, apply the boundary conditions to solve for the constants and Thus resulting in the solution,

Solve the Boundary Value Problem (BVP).

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is, Therefore, the equation becomes From here, apply the boundary conditions to solve for the constants and Thus resulting in the solution,

Solve the Boundary Value Problem (BVP).

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is, Therefore, the equation becomes From here, apply the boundary conditions to solve for the constants and Thus resulting in the solution,

Solve the Boundary Value Problem (BVP).

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is, Therefore, the equation becomes From here, apply the boundary conditions to solve for the constants and Thus resulting in the solution,

Solve the Boundary Value Problem (BVP).

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is, Therefore, the equation becomes From here, apply the boundary conditions to solve for the constants and Thus resulting in the solution,

Solve the Boundary Value Problem (BVP).

To solve this Boundary Value Problem (BVP) recall that the general solution for this type of derivative is, Therefore, the equation becomes From here, apply the boundary conditions to solve for the constants and Thus resulting in the solution,

Solve the Boundary Value Problem (BVP).

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

Solve the Boundary Value Problem (BVP).

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

Solve the Boundary Value Problem (BVP).

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

Solve the Boundary Value Problem (BVP).

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

Solve the Boundary Value Problem (BVP).

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

Solve the Boundary Value Problem (BVP).

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

Solve the Boundary Value Problem (BVP).

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

Solve the Boundary Value Problem (BVP).

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