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Q1. Solve the differential equation:

Solution

Q2.

Let space straight y left parenthesis straight x right parenthesis space be space the space solution space of space the space differential space equation space open parentheses straight x space log space straight x close parentheses dy over dx plus straight y equals 2 xlog space straight x left parenthesis straight x greater or equal than 1 right parenthesis Then space straight y left parenthesis straight e right parenthesis space is space equal space to space colon

1) 0
2) 2e
3) e
4) 2

Solution

Q3. Solve the differential equation:

Solution

The given differential equation is:

Q4. Solve the following differential equation:

Solution

The given differential equation is:

Q5. Solve the differential equation

Solution

Q6. Solve the differential equation

Solution

Q7. Solve:

Solution

The given differential equation is:

Q8. Form a differential equation of the equation: y² + 2ay + x² = 0.

Solution

The equation is: y² + 2ay + x² = 0 …(1) Differentiating w. r. to x, we get

rightwards double arrow y fraction numerator d y over denominator d x end fraction plus a fraction numerator d y over denominator d x end fraction plus x equals 0 rightwards double arrow a equals fraction numerator negative x minus y fraction numerator d y over denominator d x end fraction over denominator fraction numerator d y over denominator d x end fraction end fraction S u b s t i t u t i n g space t h e space v a l u e space o f space a space i n space t h e space o r i g i n a l space e q u a t i o n space w e space g e t y squared plus 2 y open parentheses fraction numerator negative x minus y fraction numerator d y over denominator d x end fraction over denominator fraction numerator d y over denominator d x end fraction end fraction close parentheses plus x squared equals 0 y squared fraction numerator d y over denominator d x end fraction minus 2 x y minus 2 y squared fraction numerator d y over denominator d x end fraction plus x squared fraction numerator d y over denominator d x end fraction equals 0 left parenthesis x squared minus y squared right parenthesis fraction numerator d y over denominator d x end fraction minus 2 x y equals 0

Q9. Solve the differential equation

Solution

Q10. Solve the equation:

Solution

The given differential equation is:

Q11. Show that the function f, defined by

satisfies the initial value problems y’ + y = 2, y(0) = 3.

Solution

Here, Differentiate w. r. t x, we get If we put y for f(x), then So, f satisfies the initial condition.

Q12. Solve the differential equation

Solution

The given differential equation is:

Q13. Solve the differential equation

Solution

The given differential equation is:

Q14. Show that

is a solution of differential equation

Solution

Here, we have the equation. Differentiate w. r. to x, we get _{$fraction numerator d y over denominator d x end fraction equals negative a over x sin left parenthesis log x right parenthesis plus b over x cos left parenthesis log x right parenthesis rightwards double arrow x fraction numerator d y over denominator d x end fraction equals negative a sin left parenthesis log x right parenthesis plus b cos left parenthesis log x right parenthesis A g a i n space d i f f e r e n t i a t i n g space w. r. t space x space w e space g e t x fraction numerator d squared y over denominator d x squared end fraction plus fraction numerator d y over denominator d x end fraction equals negative a over x cos left parenthesis log x right parenthesis minus b over x sin left parenthesis log x right parenthesis rightwards double arrow x squared fraction numerator d squared y over denominator d x squared end fraction plus x fraction numerator d y over denominator d x end fraction equals negative left parenthesis a cos left parenthesis log x right parenthesis plus b sin left parenthesis log x right parenthesis right parenthesis rightwards double arrow x squared fraction numerator d squared y over denominator d x squared end fraction plus x fraction numerator d y over denominator d x end fraction equals negative y rightwards double arrow x squared fraction numerator d squared y over denominator d x squared end fraction plus x fraction numerator d y over denominator d x end fraction plus y equals 0$} Which is the required differential equation

Q15. Find the order and degree of the differential equation:

Solution

Since, the highest differential coefficient of the equation is and power of is one. Therefore, order of the equation is 2 and degree is 1.

Q16. Form the differential equation of the family of curves:

Solution

The equation of the family of curves is Since, the equation (1) has two arbitrary constants a and b, so we differentiate it twice. Which is the required differential equation.