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Differential Equations

In this chapter, we will shidy about differential equations and their solutions. Concepts of differential equations will help us in this class and also in higher studies. It has many applications in other branches like science, algebra, etc. Key topics of this chapter are - definition, order and degree, general and particular solutions, formation of differential equaĊˆon whose general solution is given, method of separation of variables, homogeneous differential equations of first order and degree.

Download pdf of NCERT Solutions for Class mathematics Chapter 9 Differential Equations

Download pdf of NCERT Examplar with Solutions for Class mathematics Chapter 9 Differential Equations

Exercise 1

Exercise 2

  • Q1

    y = ex +1 : yn -y' = 0

    Ans:

    y = ex +1

    Differentiating both sides of this equation with respect to x, we get:

    \begin{align}\frac{dy}{dx}=\frac{d}{dx}(e^x + 1)\end{align}

    => y' = ex                          ...(1)

    Now, differentiating equation (1) with respect to x, we get:

    \begin{align}\frac{d}{dx}(y^{'})=\frac{d}{dx}(e^x)\end{align}

    => y'' = ex

    Substituting the values of y' and y'' in the given differential equation, we get the L.H.S. as:

    y'' - y' = ex - ex = 0 = R.H.S.

    Thus, the given function is the solution of the corresponding differential equation.


    Q2

    y = x2 + 2x + C y' - 2x - 2 = 0   

    Ans:

    y = x2 + 2x + C

    Differentiating both sides of this equation with respect to x, we get:

    \begin{align}y^{'}=\frac{d}{dx}(x^2 + 2x + C)\end{align}

    => y' = 2x + 2

    Substituting the value of y' in the given differential equation, we get:

    L.H.S. = y' - 2x - 2 = 2x + 2 - 2x - 2 = 0 = R.H.S.

    Hence, the given function is the solution of the corresponding differential equation.


    Q3

    y = cosx + C y' + sinx = 0 

    Ans:

    y = cosx + C

    Differentiating both sides of this equation with respect to x, we get:

    \begin{align}y^{'}=\frac{d}{dx}(cosx + C)\end{align}

    => y' = - sinx

    Substituting the value of y'  in the given differential equation, we get:

    L.H.S. = y' + sinx = - sinx + sinx = 0 = R.H.S.

    Hence, the given function is the solution of the corresponding differential equation.


    Q4

    \begin{align} y= \sqrt{1+x^2} : y^{'}=\frac{xy}{1+x^2}\end{align}

    Ans:

    \begin{align} y= \sqrt{1+x^2}\end{align}

    Differentiating both sides of the equation with respect to x, we get:

    \begin{align}  y^{'}=\frac{d}{dx}\left(\sqrt{1+x^2} \right)\end{align}

    \begin{align}  y^{'}=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}\left(1+x^2\right)\end{align}

    \begin{align}  y^{'}=\frac{2x}{2\sqrt{1+x^2}}\end{align}

    \begin{align}  y^{'}=\frac{x}{\sqrt{1+x^2}}\end{align}

    \begin{align}\Rightarrow y^{'}=\frac{x}{\sqrt{1+x^2}}\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\end{align}

    \begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{\sqrt{1+x^2}}\end{align}

    \begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{y}\end{align}

    \begin{align}\Rightarrow y^{'}=\frac{xy}{1+x^2}\end{align}

    ∴ L.H.S. = R.H.S.

    Hence, the given function is the solution of the corresponding differential equation.

     

     


    Q5

    y = Ax : xy' = y (x  0)

    Ans:

    y = Ax

    Differentiating both sides with respect to x, we get: 

    \begin{align}y^{'}=\frac{d}{dx}(Ax)\end{align}

    ⇒ y ' = A

    Substituting the value of y' in the given differential equation, we get:

    L.H.S. = xy= xA = Ax = y = R.H.S.

    Hence, the given function is the solution of the corresponding differential equation.


    Q6

    \begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}

    Ans:

    y= x.sinx

    Differentiating both sides of this equation with respect to x, we get:

    \begin{align} y^{'} =\frac{d}{dx}\left(x.sinx\right)\end{align}

    \begin{align}\Rightarrow y^{'} =sinx. \frac{d}{dx}\left(x\right)+ x. \frac{d}{dx}\left(sinx\right)\end{align}

    \begin{align} \Rightarrow y^{'} =sinx + x.cosx\end{align}

    Differentiating both sides of this equation with respect to x, we get:

    L.H.S. =xy= x(sinx + xcosx)

    \begin{align} =x.sinx + x^2.cosx\end{align}

    \begin{align} =y + x^2.\sqrt{1-sin^2x}\end{align}

    \begin{align} =y + x^2.\sqrt{1-\left(\frac{y}{x}\right)^2}\end{align}

    \begin{align} =y + x^2.\sqrt{\frac{x^2-y^2}{x^2}}\end{align}

    \begin{align} =y + x.\sqrt{x^2-y^2}\end{align}

    R.H.S.


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