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Q.1 Determine whether y = (c - e ^ x)/(2x); y^ prime =- 2y+e^ x 2x is a solution for the differential equation Q.2 Solve the Initial value problem ln(y ^ x) * (dy)/(dx) = 3x ^ 2 * y given y(2) = e ^ 3 . Q.3 Find the general solution for the given differential equation. (dy)/(dx) = (2x - y)/(x - 2y)

Q1 Determine Whether Y C E X2x Y Prime 2ye X 2x Is A Solution For The Differential Equation Q2 Solve The Initial Value Problem Lny X Dydx 3x 2 Y Given Y2 E 3 Q3 class=

Sagot :

LammettHash

(Q.1)

[tex]y = \dfrac{C - e^x}{2x} \implies y' = \dfrac{-2xe^x-2C+2e^x}{4x^2} = \dfrac{-xe^x-C+e^x}{2x^2}[/tex]

Then substituting into the DE gives

[tex]\dfrac{-xe^x-C+e^x}{2x^2} = -\dfrac{2\left(\dfrac{C-e^x}{2x}\right) + e^x}{2x}[/tex]

[tex]\dfrac{-xe^x-C+e^x}{2x^2} = -\dfrac{C-e^x + xe^x}{2x^2}[/tex]

[tex]\dfrac{-xe^x-C+e^x}{2x^2} = \dfrac{-C+e^x - xe^x}{2x^2}[/tex]

and both sides match, so y is indeed a valid solution.

(Q.2)

[tex]\ln\left(y^x\right)\dfrac{\mathrm dy}{\mathrm dx} = 3x^2y[/tex]

This DE is separable, since you can write [tex]\ln\left(y^x\right)=x\ln(y)[/tex]. So you have

[tex]x\ln(y)\dfrac{\mathrm dy}{\mathrm dx} = 3x^2y[/tex]

[tex]\dfrac{\ln(y)}y\,\mathrm dy = 3x\,\mathrm dx[/tex]

Integrate both sides (on the left, the numerator suggests a substitution):

[tex]\dfrac12 \ln^2(y) = \dfrac32 x^2 + C[/tex]

Given y (2) = e ³, we find

[tex]\dfrac12 \ln^2(e^3) = 6 + C[/tex]

[tex]C = \dfrac12 \times3^2 - 6 = -\dfrac32[/tex]

so that the particular solution is

[tex]\dfrac12 \ln^2(y) = \dfrac32 x^2 - \dfrac32[/tex]

[tex]\ln(y) = \pm\sqrt{3x^2 - 3}[/tex]

[tex]\boxed{y = e^{\pm\sqrt{3x^2-3}}}[/tex]

(Q.3) I believe I've already covered in another question you posted.

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