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Differential Equations question

2024 · 27 Jan · Shift 1 · Q52
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  5. /2024 · 27 Jan · Shift 1 · Q52

Differential Equations question

2024 · 27 Jan · Shift 1 · Q52

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
If the solution of the differential equation (2x+3y−2)dx+(4x+6y−7)dy=0,y(0)=3(2 x+3 y-2) \mathrm{d} x+(4 x+6 y-7) \mathrm{d} y=0, y(0)=3(2x+3y−2)dx+(4x+6y−7)dy=0,y(0)=3, is αx+βy+3log⁡e∣2x+3y−γ∣=6\alpha x+\beta y+3 \log _e|2 x+3 y-\gamma|=6αx+βy+3loge​∣2x+3y−γ∣=6, then α+2β+3γ\alpha+2 \beta+3 \gammaα+2β+3γ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 29

  1. Given differential equation
(2x+3y−2) dx+(4x+6y−7) dy=0,y(0)=3(2x+3y-2)\,dx+(4x+6y-7)\,dy=0, \qquad y(0)=3(2x+3y−2)dx+(4x+6y−7)dy=0,y(0)=3

Let

M=2x+3y−2,N=4x+6y−7.M=2x+3y-2, \qquad N=4x+6y-7.M=2x+3y−2,N=4x+6y−7.

Then the equation is

M dx+N dy=0.M\,dx+N\,dy=0.Mdx+Ndy=0.
  1. Notice a useful substitution

Both MMM and NNN involve the combination 2x+3y2x+3y2x+3y. So set

u=2x+3y.u=2x+3y.u=2x+3y.

Then

dνdx=2+3dydx.\frac{d\nu}{dx}=2+3\frac{dy}{dx}.dxdν​=2+3dxdy​.

From the differential equation,

(2x+3y−2)+(4x+6y−7)dydx=0.(2x+3y-2)+(4x+6y-7)\frac{dy}{dx}=0.(2x+3y−2)+(4x+6y−7)dxdy​=0.

Since 2x+3y=ν2x+3y=\nu2x+3y=ν, we get

(ν−2)+(2ν−7)dydx=0.(\nu-2)+(2\nu-7)\frac{dy}{dx}=0.(ν−2)+(2ν−7)dxdy​=0.

So

dydx=−ν−22ν−7.\frac{dy}{dx}=-\frac{\nu-2}{2\nu-7}.dxdy​=−2ν−7ν−2​.

Hence

dνdx=2+3(−ν−22ν−7)=2(2ν−7)−3(ν−2)2ν−7=4ν−14−3ν+62ν−7=ν−82ν−7.\frac{d\nu}{dx}=2+3\left(-\frac{\nu-2}{2\nu-7}\right) =\frac{2(2\nu-7)-3(\nu-2)}{2\nu-7} =\frac{4\nu-14-3\nu+6}{2\nu-7} =\frac{\nu-8}{2\nu-7}.dxdν​=2+3(−2ν−7ν−2​)=2ν−72(2ν−7)−3(ν−2)​=2ν−74ν−14−3ν+6​=2ν−7ν−8​.

Therefore,

dxdν=2ν−7ν−8.\frac{dx}{d\nu}=\frac{2\nu-7}{\nu-8}.dνdx​=ν−82ν−7​.
  1. Integrate

Rewrite:

2ν−7ν−8=2+9ν−8.\frac{2\nu-7}{\nu-8}=2+\frac{9}{\nu-8}.ν−82ν−7​=2+ν−89​.

So

dx=(2+9ν−8)dν.dx=\left(2+\frac{9}{\nu-8}\right)d\nu.dx=(2+ν−89​)dν.

Integrating,

x=2ν+9ln⁡∣ν−8∣+C.x=2\nu+9\ln|\nu-8|+C.x=2ν+9ln∣ν−8∣+C.

Bring all terms to one side:

2ν−x+9ln⁡∣ν−8∣=C1.2\nu-x+9\ln|\nu-8|=C_1.2ν−x+9ln∣ν−8∣=C1​.

Now substitute back ν=2x+3y\nu=2x+3yν=2x+3y:

2(2x+3y)−x+9ln⁡∣2x+3y−8∣=C1,2(2x+3y)-x+9\ln|2x+3y-8|=C_1,2(2x+3y)−x+9ln∣2x+3y−8∣=C1​,

which gives

3x+6y+9ln⁡∣2x+3y−8∣=C1.3x+6y+9\ln|2x+3y-8|=C_1.3x+6y+9ln∣2x+3y−8∣=C1​.

Dividing by 333,

x+2y+3ln⁡∣2x+3y−8∣=C2.x+2y+3\ln|2x+3y-8|=C_2.x+2y+3ln∣2x+3y−8∣=C2​.
  1. Use the initial condition

Given y(0)=3y(0)=3y(0)=3:

0+2(3)+3ln⁡∣2(0)+3(3)−8∣=C2.0+2(3)+3\ln|2(0)+3(3)-8|=C_2.0+2(3)+3ln∣2(0)+3(3)−8∣=C2​.

So

6+3ln⁡∣1∣=C2=6.6+3\ln|1|=C_2=6.6+3ln∣1∣=C2​=6.

Thus the solution is

x+2y+3ln⁡∣2x+3y−8∣=6.x+2y+3\ln|2x+3y-8|=6.x+2y+3ln∣2x+3y−8∣=6.

Comparing with

αx+βy+3log⁡e∣2x+3y−γ∣=6,\alpha x+\beta y+3\log_e|2x+3y-\gamma|=6,αx+βy+3loge​∣2x+3y−γ∣=6,

we get

α=1,β=2,γ=8.\alpha=1,\qquad \beta=2,\qquad \gamma=8.α=1,β=2,γ=8.
  1. Compute the required value
α+2β+3γ=1+2(2)+3(8)=1+4+24=29.\alpha+2\beta+3\gamma=1+2(2)+3(8)=1+4+24=29.α+2β+3γ=1+2(2)+3(8)=1+4+24=29.
  1. Comparison with stored answer

Stored correct answer: 292929

Our derived answer is also 292929, so they agree.

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