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Miscellaneous Examples · Example 20

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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Exponentiating turns the equation into a separable one; with y(0)=0y(0)=0 the particular solution is 4e3x+3e−4y=74e^{3x}+3e^{-4y}=7.

Remove the logarithm

log⁡ ⁣(dydx)=3x+4y ⇒ dydx=e3x+4y=e3xe4y.\log\!\left(\frac{dy}{dx}\right)=3x+4y\ \Rightarrow\ \frac{dy}{dx}=e^{3x+4y}=e^{3x}e^{4y}.

Now the right side is a product of a function of xx and a function of yy, so it separates.

Separate

Divide by e4ye^{4y} (never zero) and multiply by dxdx:

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

Integrate

∫e−4y dy=∫e3x dx ⇒ −14e−4y=13e3x+C.\int e^{-4y}\,dy=\int e^{3x}\,dx\ \Rightarrow\ -\tfrac14 e^{-4y}=\tfrac13 e^{3x}+C.

Apply the initial condition

At x=0x=0, y=0y=0:

−14e0=13e0+C ⇒ −14=13+C ⇒ C=−712.-\tfrac14 e^{0}=\tfrac13 e^{0}+C\ \Rightarrow\ -\tfrac14=\tfrac13+C\ \Rightarrow\ C=-\tfrac{7}{12}.

Clean up

Multiply −14e−4y=13e3x−712-\tfrac14 e^{-4y}=\tfrac13 e^{3x}-\tfrac{7}{12} by −12-12:

3e−4y=−4e3x+7 ⇒ 4e3x+3e−4y=7.3e^{-4y}=-4e^{3x}+7\ \Rightarrow\ 4e^{3x}+3e^{-4y}=7.

Check: differentiating gives 12e3x−12e−4y y′=012e^{3x}-12e^{-4y}\,y'=0, so y′=e3xe4yy'=e^{3x}e^{4y} and log⁡y′=3x+4y\log y'=3x+4y; also 4+3=74+3=7 at the origin. ✓

✓Final answer

4e3x+3e−4y=74e^{3x}+3e^{-4y}=7.

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