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Exercise 4 · Q8

Q.Find the particular solution of the differential equation log⁡(dydx)=3x+4y\log\left(\frac{dy}{dx}\right)=3x+4y, given that y=0y=0, when x=0x=0.

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Rewrite log⁡ ⁣(dydx)=3x+4y\log\!\big(\tfrac{dy}{dx}\big)=3x+4y as dydx=e3xe4y\tfrac{dy}{dx}=e^{3x}e^{4y}, separate, integrate, and use (0,0)(0,0) to get 4e3x+3e−4y=74e^{3x}+3e^{-4y}=7.

log⁡(⋅)\log(\cdot) means natural log, so dydx=e3x+4y=e3x e4y\dfrac{dy}{dx}=e^{3x+4y}=e^{3x}\,e^{4y} — variables separable.

Steps

  1. Given:

log⁡(dydx)=3x+4y,y=0 at x=0.\log\left(\frac{dy}{dx}\right)=3x+4y,\quad y=0\text{ at }x=0.

  1. Exponentiate:

dydx=e3x+4y=e3x e4y.\frac{dy}{dx}=e^{3x+4y}=e^{3x}\,e^{4y}.

  1. Separate variables:

e−4y dy=e3x dx.e^{-4y}\,dy=e^{3x}\,dx.

  1. Integrate both sides:

−14e−4y=13e3x+C.(1)-\frac14 e^{-4y}=\frac13 e^{3x}+C.\qquad(1)

  1. Apply x=0, y=0x=0,\ y=0 (e0=1e^{0}=1): …

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