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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the particular solution of the differential equation $ log \bigg( \large\frac{dy}{dx} \bigg) =3x+4y,$ given that $ y=0$ when $x=0$

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dy/dx=e^(3x+4y)

Let 3x+4y=z

⇒3+4 dy/dx=dz/dx

⇒dy/dx=(dz/dx-3)/4

Substituting the value of dy/dx

(dz/dx-3)/4=e^z

⇒dz/dx=4e^z+3

⇒dz/(4e^z+3)=dx

IIntegrating both the sides

Int (  e^z.dz)/(4(e^z)^2+3e^z)=Int dx

Put e^z=t  ⇒e^z.dz=dt

Int   dt/(4t^2+3t)=x+c

1/4.  Int   dt/(t+3/8)^2-9/64)=x+c

⇒ 1/4  4/3  log(4t-3)/(4t+3)=x+c

⇒1/3  log(4e^3x+4y-3)/(4e^3x+4y+3)=x+c

Given x=0  and y=0

⇒c=1/3  log(1/7)

Substitute the value of c  and the particular solution is

1/3 { log7(4e^3x+4y-3)/(4e^3x+4y+3)}=x
answered May 17, 2015 by rvidyagovindarajan_1
 

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