Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

Find `A`and `B`so that `y=Asin3x+bcos3x`satisfies the equation`(d^2y)/(dx^2)+4(dy)/(dx)+3y=10cos3xdot`

Answer» `y=Asin3x+bcos3x`
diff. with respect to x
`dy/dx=Acos3x*3+B(-Sin3x)*3`
diff with respect to x
`(d^2y)/(dx^2)=-9Asin3x-9Bcos3x`
`(d^2y)/(dx^2)+4dy/dx+3y=10`
LHS`=-9Asin3x-9Bcos3x+12Acos3x-12Bsin3x+3Asin3x+3Bcos3x=10cos3x`
`sin3x(-9A-12B+3A)+cos3x(-9B+12A+3B)=10cos3x`
`12A-6B=10`
`12*(-2B)-6B=10`
`B=-1/3`
`A=2/3`.
Discussion
2.

If`y=(sin^(-1)x)^2` ,prove that `(1-x^2)y_2-x y_1-2=0.`

Answer» `y_1=dy/dx=2*sin^(-1)x*1/sqrt(1-x^2)=(2sin^(-1)x)/sqrt(1-x^2)`
`y_1=(d^2y)/(dx^2)=2[1/(1-x^2)-1/2sin^(-1)x(1-x^2)^(-3/2)*(-2)]/(1-x^2)`
`y_2=2/(1-x^2)+(2sin^(-1)x)/(1-x^2)^(3/2)`
`(1-x^2)y_2-xy_1`
`2+(2sin^(-1)x)/(1-x^2)^(1/2)-(2sin^(-1)x)/(1-x^2)^(1/2)=2`
Discussion
3.

If`x=asint-bcost ,y=acost+bsint ,"p r o v et h a t"``(d^2y)/(dx^2)=-(x^2+y^2)/(y^3)`

Answer» Here, `x = asint - bcost` and `y = acost+bsint`
`:. dy/dx = (dy/dt)/(dx/dt) = (-asint+bcost)/(acost+bsint)`
Now, `(d^2y)/dx^2 = (d/dt(dy/dx))/(dx/dt) `
`=((-acost-bsint)(acost+bsint)-(-asint+bcost)(-asint+bcost))/((acost+bsint)^2(acost+bsint)`
`=(-(acost+bsint)^2 - (bcost-asint)^2)/((acost+bsint)^3)`
`=(-x^2-y^2)/(y^3)`
`:. (d^2y)/dx^2 = (-(x^2+y^2))/(y^3).`
Discussion
4.

If `y=e^-x cos x`, show that `(d^2y)/(dx^2)= 2e^-1 sin x`

Answer» `y=e^(-x)cosx`
diff. with respect to x
`dy/dx=-e^(-x)cosx-e^x sinx`
diff. with respect to x
`(d^2y)/(dx^2)=-[-e^(-x)cosx+e^(-x)(-sinx)]-[-e^(-x)sinx+e^(-x)cosx]`
`=e^(-x)cosxx+e^(-x)sinx+e^(-x)sinxx-e^(-x)cosx`
`=2e^(-x)sinx`.
Discussion