Question 12 Marks
Solve the following differential equations:
View full question & answer→$\log \left(\frac{d y}{d x}\right)=2 x+3 y$
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$\log \left(\frac{d y}{d x}\right)=2 x+3 y$
$(y-a)^2=b(x+4)$
$y^2=a(b-x)(b+x)$
$x^2=2 y^2 \log y, x^2+y^2=x y \frac{d x}{d y}$
$x^2+y^2=r^2 ; x \frac{d y}{d x}+r \sqrt{1+\left(\frac{d y}{d x}\right)^2}=y$
$\frac{d y}{d x}+\frac{y}{x}=x^3-3$
$\cos ^2(x-2 y)=1-2 \frac{d y}{d x}$
$x+y \frac{d y}{d x}=\sec \left(x^2+y^2\right)$
$\cos \left(\frac{d y}{d x}\right)=a, a \in R, y(0)=2$
$\left(e^y+1\right) \cos x + e ^y \sin x \frac{d y}{d x}=0$, when $x =\frac{\pi}{6}, y =0$
$3 e^x \tan y d x+\left(1+e^x\right) \sec ^2 y d y=0$, when $x=0, y=\pi$
$\frac{d y}{d x}= e ^{ x + y }+ x ^2 e ^{ y }$
$2 e^{x+2 y} d x-3 d y=0$
$y^3-\frac{d y}{d x}=x^2 \frac{d y}{d x}$
cos x . cos y dy – sin x . sin y dx = 0
$\sec ^2 x \cdot \tan y d x+\sec ^2 y \cdot \tan x d y=0$
$y = a +\frac{b}{x} ; x \frac{d^2 y}{d x^2}+2 \frac{d y}{d x}=0$
$y=x^{ m } ; x^2 \frac{d^2 y}{d x^2}-m x \frac{d y}{d x}+m y=0$
$(y-a)^2=4(x-b)$
$A x^2+B y^2=1$
$x+\frac{d^2 y}{d x^2}=\sqrt{1+\left(\frac{d^2 y}{d x^2}\right)^2}$
On squaring both sides, we get
$\left(x+\frac{d^2 y}{d x^2}\right)^2=1+\left(\frac{d^2 y}{d x^2}\right)^2$
$\therefore x^2+2 x \frac{d^2 y}{d x^2}+\left(\frac{d^2 y}{d x^2}\right)^2=1+\left(\frac{d^2 y}{d x^2}\right)^2$
$\therefore x^2+2 x \frac{d^2 y}{d x^2}-1=0$
This D.E. has highest order derivative $\frac{d^2 y}{d x^2}$ with power 1 .
∴ the given D.E. has order 2 and degree 1.
$\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20$
$\therefore\left(\frac{d^3 y}{d x^3}\right)^3 \cdot\left(\frac{d y}{d x}\right)^2=20^6$
This D.E. has highest order derivative $\frac{d^3 y}{d x^3}$ with power 3 .
∴ the given D.E. has order 3 and degree 3.
$\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}=8 \frac{d^2 y}{d x^2}$
On squaring both sides, we get
$\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=8^2 \cdot\left(\frac{d^2 y}{d x^2}\right)^2$
This D.E. has highest order derivative $\frac{d^2 y}{d x^2}$ with power 2 .
∴ the given D.E. has order 2 and degree 2.
$\left(\frac{d^2 y}{d x^2}\right)^2+\cos \left(\frac{d y}{d x}\right)=0$
This D.E. has highest order derivative $\frac{d^2 y}{d x^2}$
∴ order = 2 Since this D.E.
cannot be expressed as a polynomial in differential coefficients, the degree is not defined.
$\left(y^{\prime \prime \prime}\right) 2+3 y^{\prime \prime}+3 x y^{\prime}+5 y=0$
This can be written as:
$\left(\frac{d^3 y}{d x^3}\right)^2+3 \frac{d^2 y}{d x^2}+3 x \frac{d y}{d x}+5 y=0$
This D.E. has highest order derivative $\frac{d^3 y}{d^3}$ with power 2 .
∴ The given D.E. has order 3 and degree 2.
$\frac{d^2 y}{d t^2}+\left(\frac{d y}{d t}\right)^2+7 x+5=0$
This D.E. has highest order derivative $\frac{d^2 y}{d x^2}$ with power 1 .
∴ the given D.E. has order 2 and degree 1.
$\frac{d^2 y}{d x^2}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^3 y}{d x^3}}$
On squaring both sides, we get
$\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}+x\right)^2=1+\frac{d^3 y}{d x^3}$
This D.E. has highest order derivative $\frac{d^3 y}{d^3}$ with power 1 .
∴ the given D.E. has order 3 and degree 1.
$\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}$
$\therefore\left(\frac{d y}{d x}\right)^2=2 \sin x+3$
This D.E. has highest order derivative $\frac{d y}{d x}$ with power 2 .
∴ the given D.E. is of order 1 and degree 2.
$\sqrt[3]{1+\left(\frac{d y}{d x}\right)^2}=\frac{d^2 y}{d x^2}$
On cubing both sides, we get
$1+\left(\frac{d y}{d x}\right)^2=\left(\frac{d^2 y}{d x^2}\right)^3$
This D.E. has highest order derivative $\frac{d^2 y}{d x^2}$ with power 3 .
∴ the given D.E. is of order 2 and degree 3.
$\frac{d y}{d x^2}+X\left(\frac{d y}{d x}\right)+y=2 \sin x$
This D.E. has highest order derivative $\frac{d^2 y}{d x^2}$ with power 1 .
∴ the given D.E. is of order 2 and degree 1.